Initial Setup & Algebraic Analysis

We are tasked with factorising the following binomial expression:

$27y^3 + 125z^3$

To factorise this polynomial, we must first analyze the coefficients and the degrees of the variables to identify any underlying algebraic structures. We observe that both terms are perfect cubes.

Step 1: Expressing Terms as Perfect Cubes

We determine the cube roots of the numerical coefficients and the variables:

• For the first term: The prime factorization of $27$ is $3 \times 3 \times 3 = 3^3$. Therefore, $27y^3$ can be rewritten as $(3y)^3$ [Applying the power of a product rule: $x^n y^n = (xy)^n$].
• For the second term: The prime factorization of $125$ is $5 \times 5 \times 5 = 5^3$. Therefore, $125z^3$ can be rewritten as $(5z)^3$.

Rewriting the original expression, we get:

$(3y)^3 + (5z)^3$

Step 2: Stating the Relevant Algebraic Identity

The expression is now explicitly in the form of the sum of two cubes, $a^3 + b^3$. [Per the fundamental algebraic identity for the sum of cubes, derived from the expansion of $(a+b)^3 - 3ab(a+b)$], we know that:

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$

Step 3: Substitution and Expansion

By mapping our terms to the identity, we set $a = 3y$ and $b = 5z$. Substituting these into the right-hand side of the identity yields:

$(3y)^3 + (5z)^3 = (3y + 5z) \left[ (3y)^2 - (3y)(5z) + (5z)^2 \right]$

Step 4: Simplification of the Quadratic Factor

We must now rigorously simplify each term inside the second set of parentheses (the quadratic trinomial factor):

• Square of the first term: $(3y)^2 = 3^2 \cdot y^2 = 9y^2$
• Product of the two terms: $(3y)(5z) = (3 \cdot 5)(y \cdot z) = 15yz$
• Square of the second term: $(5z)^2 = 5^2 \cdot z^2 = 25z^2$

Replacing the unsimplified terms in our expanded equation with these calculated values, we obtain the final factorised expression:

$(3y + 5z)(9y^2 - 15yz + 25z^2)$

Final Solution: The completely factorised form of $27y^3 + 125z^3$ is $(3y + 5z)(9y^2 - 15yz + 25z^2)$.

Initial Setup & Objective

We are tasked with verifying the fundamental algebraic identity for the difference of two cubes. The identity is given as:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

To verify this rigorously, we will expand the Right-Hand Side (RHS) and demonstrate that it simplifies exactly to the Left-Hand Side (LHS). We will also provide a secondary derivation using the binomial cube expansion to establish deep theoretical consistency.

Method 1: Direct Algebraic Expansion

Step 1: Applying the Distributive Property

We begin with the expression on the Right-Hand Side (RHS):

$RHS = (x - y)(x^2 + xy + y^2)$

[Per the Distributive Law of Multiplication over Addition, $A(B + C + D) = AB + AC + AD$]. We must distribute both terms of the binomial $(x)$ and $(-y)$ across the entire trinomial:

$RHS = x(x^2 + xy + y^2) - y(x^2 + xy + y^2)$

Step 2: Term-by-Term Multiplication

Next, we distribute $x$ into the first trinomial and $-y$ into the second trinomial. [By the laws of exponents, $x^a \cdot x^b = x^{a+b}$]:

• Multiplying by $x$:
  $x \cdot x^2 = x^3$
  $x \cdot xy = x^2y$
  $x \cdot y^2 = xy^2$
• Multiplying by $-y$:
  $-y \cdot x^2 = -x^2y$ [Note: Commutativity allows us to write $-yx^2$ as $-x^2y$]
  $-y \cdot xy = -xy^2$
  $-y \cdot y^2 = -y^3$

Substituting these expanded terms back into our equation yields:

$RHS = (x^3 + x^2y + xy^2) + (-x^2y - xy^2 - y^3)$

Step 3: Aggregation and Simplification

We now drop the parentheses and group the like terms together to identify additive inverses (terms that sum to zero):

$RHS = x^3 + x^2y - x^2y + xy^2 - xy^2 - y^3$

Rearranging the terms for absolute clarity:

$RHS = x^3 + (x^2y - x^2y) + (xy^2 - xy^2) - y^3$

The terms $x^2y$ and $-x^2y$ cancel each other out. Similarly, $xy^2$ and $-xy^2$ cancel each other out:

$RHS = x^3 + 0 + 0 - y^3$

$RHS = x^3 - y^3$

Since the simplified RHS perfectly matches the LHS, the identity is verified.

Visualizing the Expansion (Area Model Matrix)

The following matrix visually maps the distribution of $(x - y)$ across $(x^2 + xy + y^2)$. The color-coded cells demonstrate how the intermediate terms cancel each other out, leaving only the cubic terms.

Method 2: Derivation from the Binomial Cube Identity

To provide exhaustive proof, we can also derive this factorization directly from the standard identity for the cube of a binomial:

$ (x - y)^3 = x^3 - y^3 - 3xy(x - y) $

Step 1: Isolate the Difference of Cubes

We want to solve for $x^3 - y^3$. By adding $3xy(x - y)$ to both sides of the equation, we get:

$ x^3 - y^3 = (x - y)^3 + 3xy(x - y) $

Step 2: Factor out the Common Binomial

Observe that both terms on the right side share a common factor of $(x - y)$. We extract this factor:

$ x^3 - y^3 = (x - y) \left[ (x - y)^2 + 3xy \right] $

Step 3: Expand and Simplify the Inner Bracket

We expand the squared binomial $(x - y)^2 = x^2 - 2xy + y^2$ inside the brackets:

$ x^3 - y^3 = (x - y) \left[ (x^2 - 2xy + y^2) + 3xy \right] $

Combine the like terms ($-2xy$ and $+3xy$):

$ -2xy + 3xy = +xy $

Substituting this back into the bracket yields the final verified identity:

$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $

Final Solution: Through both direct algebraic expansion and derivation from the binomial cube identity, it is rigorously proven that $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$. The intermediate terms $x^2y$ and $xy^2$ act as additive inverses and cancel out perfectly.

Initial Setup & Given Polynomial

We are tasked with factorising the following algebraic expression:

$P(a) = 27 - 125a^3 - 135a + 225a^2$

Step 1: Identifying the Relevant Algebraic Identity

By analyzing the polynomial, we observe the presence of perfect cubes ($27$ and $125a^3$) and alternating signs. This structural pattern strongly indicates the application of the standard cubic identity for the difference of two terms [Per the Binomial Theorem for exponent 3]:

$(x - y)^3 = x^3 - y^3 - 3x^2y + 3xy^2$

To utilize this identity, we must map the terms of our given polynomial to the terms of the expansion.

Step 2: Extracting the Base Terms (Cube Roots)

We isolate the perfect cube terms to determine our candidate values for $x$ and $y$.

• First term (Constant): The term $27$ is a perfect cube.
  $27 = (3)^3 \implies x = 3$
• Second term (Cubic): The term $125a^3$ is a perfect cube.
  $125a^3 = (5a)^3 \implies y = 5a$

Step 3: Verifying the Cross Terms

To rigorously confirm that the polynomial fits the identity $(x - y)^3$, we must evaluate the intermediate cross terms $-3x^2y$ and $+3xy^2$ using our candidate values $x = 3$ and $y = 5a$.

• Evaluating $-3x^2y$:
  $-3(3)^2(5a) = -3(9)(5a) = -27(5a) = -135a$
  [This perfectly matches the third term of the given polynomial].
• Evaluating $+3xy^2$:
  $+3(3)(5a)^2 = 9(25a^2) = 225a^2$
  [This perfectly matches the fourth term of the given polynomial].

Visual Mapping of the Algebraic Identity

The following diagram illustrates the exact one-to-one mapping between the given polynomial terms and the components of the cubic identity.

Step 4: Synthesizing the Factorised Form

Since all terms perfectly align with the expansion of $(x - y)^3$, we can rewrite the original polynomial in its expanded identity form:

$27 - 125a^3 - 135a + 225a^2 = (3)^3 - (5a)^3 - 3(3)^2(5a) + 3(3)(5a)^2$

Applying the identity $(x - y)^3$, where $x = 3$ and $y = 5a$, we condense the expression into a single perfect cube:

$= (3 - 5a)^3$

To express this as a product of its linear factors, we write the binomial three times.

Final Solution: The completely factorised form of the polynomial is $(3 - 5a)(3 - 5a)(3 - 5a)$.

Initial Setup & Expression Analysis

We are tasked with factorising the following algebraic polynomial:

$9x^2 + 6xy + y^2$

To factorise this expression efficiently, we must analyze its structural properties. The polynomial consists of three terms (a trinomial). Our primary objective is to determine if it conforms to the structure of a perfect square trinomial, which can be factorised using standard algebraic identities.

Step 1: Decomposing the Terms into Perfect Squares

We begin by examining the first and third terms to see if they can be expressed as perfect squares [Per the fundamental properties of exponents].

• First Term: The term $9x^2$ consists of a constant $9$ and a variable $x^2$. Since $3^2 = 9$, we can rewrite this term as the square of a single monomial:
  $9x^2 = (3x)^2$
• Third Term: The term $y^2$ is already a perfect square and can be explicitly written as:
  $y^2 = (y)^2$

Step 2: Verifying the Middle Term

For a trinomial to be classified as a perfect square trinomial, the middle term must be exactly twice the product of the bases of the squared terms identified in Step 1.

Let us define our base variables based on the standard identity structure:

• Let $a = 3x$
• Let $b = y$

Now, we calculate the theoretical middle term, $2ab$:

$2ab = 2 \cdot (3x) \cdot (y)$

$2ab = 6xy$

This calculated product perfectly matches the middle term of our given polynomial ($6xy$). Therefore, the expression is confirmed to be a perfect square trinomial.

Step 3: Applying the Standard Algebraic Identity

Having verified the structure, we apply the First Algebraic Identity for the square of a binomial [By the binomial expansion theorem]:

$a^2 + 2ab + b^2 = (a + b)^2$

Substituting our specific values ($a = 3x$ and $b = y$) into the identity:

$(3x)^2 + 2(3x)(y) + (y)^2 = (3x + y)^2$

Expanding the squared binomial into its linear factors yields:

$(3x + y)^2 = (3x + y)(3x + y)$

Geometric Proof & Visualization

The algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$ can be rigorously proven using area models in Euclidean geometry. Below is the geometric representation of our specific polynomial, where the total area of a square with side length $(3x + y)$ is equal to the sum of the areas of its four internal rectangular regions.

As demonstrated by the geometric model, the total area is the sum of the individual regions: $9x^2 + 3xy + 3xy + y^2 = 9x^2 + 6xy + y^2$, which perfectly corresponds to a square of side $(3x + y)$.

Final Solution: The completely factorised form of the polynomial $9x^2 + 6xy + y^2$ is $(3x + y)(3x + y)$, which is most concisely written as $(3x + y)^2$.

Step 1: Given Variables & Initial Setup

We are tasked with evaluating the sum of three cubes without performing direct cubing operations. The given mathematical expression is:

$(-12)^3 + (7)^3 + (5)^3$

Let us define the base of each cubic term as a distinct variable:

• $a = -12$
• $b = 7$
• $c = 5$

Step 2: Evaluating the Sum of the Base Variables

Before attempting to expand the expression, we must analyze the linear sum of the base variables. This is a critical diagnostic step in polynomial algebra to determine if a conditional identity applies.

$a + b + c = (-12) + 7 + 5$

$a + b + c = -12 + 12 = 0$

[Geometrically, this represents a closed vector loop in one dimension, where the displacement from the origin returns exactly to zero. See the precise vector visualization below.]

Step 3: Theoretical Justification & Algebraic Identity

We rely on the fundamental algebraic identity for the sum of three cubes:

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

[Per the Zero Product Property], if the first factor on the right-hand side, $(a + b + c)$, is equal to zero, the entire right-hand side of the equation evaluates to zero, regardless of the values of the quadratic polynomial factor. Therefore:

$a^3 + b^3 + c^3 - 3abc = (0) \times (a^2 + b^2 + c^2 - ab - bc - ca)$

$a^3 + b^3 + c^3 - 3abc = 0$

Transposing $-3abc$ to the right side yields the conditional identity:

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Step 4: Application of the Conditional Identity

Since we have rigorously proven in Step 2 that $(-12) + 7 + 5 = 0$, we can directly substitute our variables into the conditional identity:

$(-12)^3 + (7)^3 + (5)^3 = 3 \cdot (-12) \cdot (7) \cdot (5)$

Step 5: Final Arithmetic Calculation

We now perform the sequential multiplication of the terms on the right-hand side:

• First, multiply the constants: $3 \times (-12) = -36$
• Next, multiply the remaining terms: $7 \times 5 = 35$
• Finally, compute the product of the two results: $(-36) \times 35$

To calculate $(-36) \times 35$ mentally or systematically:

$(-36) \times (30 + 5) = (-36 \times 30) + (-36 \times 5)$

$= -1080 - 180 = -1260$

Final Solution: The value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$.

Step 1: Initial Expression Analysis

We are given the quadratic trinomial:

$P(y) = 4y^2 - 4y + 1$

Our objective is to factorise this expression by identifying it as a perfect square. A quadratic trinomial takes the general form $ax^2 + bx + c$. We must analyze the first and last terms to determine if they are perfect squares.

Step 2: Identifying the Appropriate Algebraic Identity

We observe that the signs of the terms alternate ($+, -, +$). This suggests the use of the standard binomial square identity for subtraction [Per the fundamental axioms of polynomial expansion]:

$(a - b)^2 = a^2 - 2ab + b^2$

Step 3: Term-by-Term Transformation & Verification

We must map the terms of our given polynomial to the components of the identity $a^2 - 2ab + b^2$.

• First Term ($a^2$): The term $4y^2$ can be rewritten as the square of a single monomial.
  $4y^2 = (2y)^2 \implies a = 2y$
• Third Term ($b^2$): The constant $1$ is a perfect square.
  $1 = (1)^2 \implies b = 1$
• Middle Term ($-2ab$): We must verify that the middle term of our polynomial matches $-2ab$ using our derived values for $a$ and $b$.
  $-2ab = -2(2y)(1) = -4y$

Since the middle term perfectly matches the given expression ($-4y$), the polynomial is confirmed to be a perfect square trinomial.

Step 4: Applying the Identity

By substituting $a = 2y$ and $b = 1$ back into the structural framework of the identity, we can rewrite the expanded polynomial as a squared binomial:

$4y^2 - 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2$

$4y^2 - 4y + 1 = (2y - 1)^2$

Step 5: Geometric Verification (Area Model)

To rigorously prove this factorization, we can use an area model. Consider a large square with a total side length of $2y$. Its total area is $(2y)^2 = 4y^2$. If we partition this square into segments of length $(2y - 1)$ and $1$, we can geometrically derive the identity.

From the geometric model, the total area of the square is the sum of its four internal regions:

$Area_{Total} = (2y - 1)^2 + (2y - 1) + (2y - 1) + 1$

$4y^2 = (2y - 1)^2 + 4y - 2 + 1$

$4y^2 = (2y - 1)^2 + 4y - 1$

Isolating the primary region $(2y - 1)^2$ yields:

$(2y - 1)^2 = 4y^2 - 4y + 1$

[This confirms our algebraic factorization is geometrically absolute].

Step 6: Final Factorisation

To express the polynomial fully in its factorised form, we write the squared binomial as the product of two identical linear binomials.

$(2y - 1)^2 = (2y - 1)(2y - 1)$

Final Solution: The factorised form of the polynomial $4y^2 - 4y + 1$ is $(2y - 1)(2y - 1)$.

Given Expression & Algebraic Identity

We are tasked with expanding the following trinomial squared:

$(3a - 7b - c)^2$

To expand this expression systematically, we utilize the standard algebraic identity for the square of a trinomial [Derived from the distributive property of multiplication over addition]:

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

Step 1: Variable Mapping

We must map the terms of our specific expression to the variables in the standard identity. By rewriting the given expression as a sum, we ensure that the negative signs are correctly associated with their respective terms:

$(3a - 7b - c)^2 = (3a + (-7b) + (-c))^2$

By direct comparison with $(x + y + z)^2$, we establish the following variable assignments:

• $x = 3a$
• $y = -7b$
• $z = -c$

Step 2: Substitution into the Identity

Substituting these mapped values into the expanded form of the identity yields:

$(3a + (-7b) + (-c))^2 = (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$

Step 3: Term-by-Term Simplification

We now evaluate each term individually, applying the rules of exponents [$(ab)^n = a^n b^n$] and the rules of sign multiplication:

• Squaring the first term: $(3a)^2 = 3^2 \cdot a^2 = 9a^2$
• Squaring the second term: $(-7b)^2 = (-7)^2 \cdot b^2 = 49b^2$ [Note: The square of a negative number is positive]
• Squaring the third term: $(-c)^2 = (-1)^2 \cdot c^2 = c^2$
• First cross-product: $2(3a)(-7b) = 2 \cdot 3 \cdot (-7) \cdot a \cdot b = -42ab$
• Second cross-product: $2(-7b)(-c) = 2 \cdot (-7) \cdot (-1) \cdot b \cdot c = 14bc$ [Note: The product of two negative numbers is positive]
• Third cross-product: $2(-c)(3a) = 2 \cdot (-1) \cdot 3 \cdot c \cdot a = -6ca$

Step 4: Assembling the Final Polynomial

Combining all the simplified terms from Step 3, we construct the final expanded polynomial:

$9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$

Geometric Representation of the Trinomial Square Identity

The algebraic identity $(x+y+z)^2$ can be visualized as the area of a square with side length $(x+y+z)$, partitioned into 9 distinct rectangular regions. The sum of the areas of these regions corresponds exactly to the terms in our expanded formula.

Final Solution: $9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca$

Step 1: Initial Setup & Identification of the Algebraic Identity

We are tasked with finding the product of the binomials $(3 - 2x)$ and $(3 + 2x)$.

By analyzing the structure of the expression, we observe that it consists of the product of the difference and the sum of the exact same two terms. This perfectly matches the standard algebraic identity for the Difference of Squares.

The Difference of Squares identity is defined as:

$(a - b)(a + b) = a^2 - b^2$

[Theoretical Justification: This identity is derived via the distributive property (FOIL method), where $(a - b)(a + b) = a^2 + ab - ab - b^2$. The middle terms $+ab$ and $-ab$ cancel each other out, leaving $a^2 - b^2$.]

Step 2: Variable Mapping

To apply the identity, we map the terms from our specific expression $(3 - 2x)(3 + 2x)$ to the general variables $a$ and $b$ in the identity:

• Let $a = 3$
• Let $b = 2x$

Step 3: Application of the Identity

Substituting the mapped variables into the right-hand side of the identity $(a^2 - b^2)$, we construct the following equation:

$(3 - 2x)(3 + 2x) = (3)^2 - (2x)^2$

Step 4: Algebraic Simplification

Now, we evaluate the squares for both terms independently:

• First term: $(3)^2 = 3 \times 3 = 9$
• Second term: $(2x)^2 = (2)^2 \cdot (x)^2 = 4x^2$
  [Per the Power of a Product Rule in exponentiation: $(xy)^n = x^n y^n$]

Substituting these evaluated squares back into our equation yields:

$(3)^2 - (2x)^2 = 9 - 4x^2$

Geometric Representation of the Difference of Squares

The algebraic identity $(a - b)(a + b) = a^2 - b^2$ can be visualized geometrically. If we take a large square of area $a^2$ and remove a smaller square of area $b^2$, the remaining area can be rearranged into a rectangle with dimensions $(a - b)$ and $(a + b)$.

Final Solution: The product of $(3 - 2x)(3 + 2x)$ is $9 - 4x^2$.

Initial Setup & Algebraic Objective

We are required to rigorously verify the standard algebraic identity for the sum of two cubes. The equation provided is:

$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$

To prove this equivalence, we will designate the expressions as follows:

• Left Hand Side (LHS): $x^3 + y^3$
• Right Hand Side (RHS): $(x + y)(x^2 - xy + y^2)$

Our objective is to expand the RHS and demonstrate that it simplifies exactly to the LHS.

Step 1: Applying the Distributive Property

We begin by expanding the RHS using the distributive property of multiplication over addition [Algebraic Axiom: $a(b + c + d) = ab + ac + ad$]. We must distribute both terms of the binomial $(x + y)$ across the trinomial $(x^2 - xy + y^2)$.

$RHS = x(x^2 - xy + y^2) + y(x^2 - xy + y^2)$

Step 2: Term-by-Term Algebraic Expansion

Next, we perform the scalar multiplication for each distributed component, applying the laws of exponents [Specifically, the Product Rule: $a^m \cdot a^n = a^{m+n}$].

Expanding the first cluster:

• $x \cdot (x^2) = x^3$
• $x \cdot (-xy) = -x^2y$
• $x \cdot (y^2) = xy^2$

Resulting polynomial segment: $x^3 - x^2y + xy^2$

Expanding the second cluster:

• $y \cdot (x^2) = x^2y$
• $y \cdot (-xy) = -xy^2$
• $y \cdot (y^2) = y^3$

Resulting polynomial segment: $x^2y - xy^2 + y^3$

Step 3: Aggregating and Simplifying Like Terms

We now combine both expanded segments into a single, unified polynomial expression:

$RHS = x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$

To simplify, we group the "like terms" (terms containing the exact same variables raised to the exact same powers) using the Commutative Property of Addition:

$RHS = x^3 + (-x^2y + x^2y) + (xy^2 - xy^2) + y^3$

Because these pairs are additive inverses of one another, they sum to zero and cancel out completely.

Step 4: Final Equivalence

Substituting the zeroed terms back into the equation yields:

$RHS = x^3 + 0 + 0 + y^3$

$RHS = x^3 + y^3$

Comparing this result to our initial setup, we observe that the simplified RHS is identical to the LHS.

$LHS = RHS$

Final Solution: The identity $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$ is rigorously verified.

Given Variables & Initial Setup

We are given the linear equation involving three variables:

$x + y + z = 0$

Objective: Prove the algebraic relationship $x^3 + y^3 + z^3 = 3xyz$.

To provide a comprehensive, masterclass-level proof, this relationship will be demonstrated using two distinct mathematical approaches: the Standard Identity Method and the Direct Algebraic Manipulation (Cubing) Method.

Method 1: Utilizing the Standard Algebraic Identity

Step 1: State the relevant polynomial identity

In algebra, the sum of three cubes is governed by the following fundamental identity:

$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

[Theoretical Justification: This identity is derived by expanding the right-hand side and canceling out the intermediate cross-terms, leaving only the sum of the cubes and the $-3xyz$ term.]

Step 2: Substitute the given condition

We are given the premise that $x + y + z = 0$. We substitute this value directly into the right-hand side of our identity:

$x^3 + y^3 + z^3 - 3xyz = (0) \cdot (x^2 + y^2 + z^2 - xy - yz - zx)$

Step 3: Apply the Zero Product Property

According to the Zero Product Property, any finite real number or algebraic expression multiplied by zero results in zero. Therefore, the entire right-hand side collapses to $0$:

$x^3 + y^3 + z^3 - 3xyz = 0$

Step 4: Isolate the sum of the cubes

By adding $3xyz$ to both sides of the equation, we arrive at the final required expression:

$x^3 + y^3 + z^3 = 3xyz$

Method 2: Direct Algebraic Manipulation (Cubing)

Step 1: Rearrange the initial equation

Starting with the given condition, isolate two variables on one side of the equation:

$x + y + z = 0 \implies x + y = -z$

Step 2: Cube both sides of the equation

To generate the cubic terms required for the proof, apply the power of 3 to both sides:

$(x + y)^3 = (-z)^3$

Step 3: Expand using the binomial cube identity

Expand the left side using the standard binomial identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$. Note that the cube of a negative value remains negative, so $(-z)^3 = -z^3$.

$x^3 + y^3 + 3xy(x + y) = -z^3$

Step 4: Substitute the initial rearranged condition

From Step 1, we established that $(x + y) = -z$. Substitute $-z$ back into the expanded equation in place of $(x + y)$:

$x^3 + y^3 + 3xy(-z) = -z^3$

Step 5: Simplify and rearrange terms

Multiply the terms to simplify the equation:

$x^3 + y^3 - 3xyz = -z^3$

Finally, transpose $-z^3$ to the left side (becoming $+z^3$) and $-3xyz$ to the right side (becoming $+3xyz$):

$x^3 + y^3 + z^3 = 3xyz$

Visual Representation of the Logical Pathways

The following flowchart illustrates the dual algebraic pathways utilized to prove the theorem, confirming the structural integrity of both methods.

Final Conclusion

Both the application of the standard cubic polynomial identity and direct algebraic expansion yield the exact same mathematical truth. When the sum of three variables is zero, the sum of their cubes is perfectly balanced by three times their product.

Final Solution: It is proven that if $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.

Step 1: Initial Setup & Identification of the Algebraic Identity

We are tasked with finding the product of the given binomials:

$ \left(y^2 + \frac{3}{2}\right) \left(y^2 - \frac{3}{2}\right) $

By analyzing the structural form of the expression, we observe that it consists of the product of the sum and difference of the exact same two terms. This perfectly matches the fundamental algebraic identity for the Difference of Squares:

$ (a + b)(a - b) = a^2 - b^2 $

[Theoretical Justification: The cross-terms in the expansion $(a)(a) - (a)(b) + (b)(a) - (b)(b)$ cancel out, leaving only the squared terms $a^2 - b^2$.]

Step 2: Variable Mapping and Substitution

To apply the identity rigorously, we establish a one-to-one correspondence between the variables in the identity and the terms in our specific expression.

• Let the first term $a = y^2$
• Let the second term $b = \frac{3}{2}$

Substituting these defined values into the right-hand side of the Difference of Squares identity ($a^2 - b^2$), we formulate the following equation:

$ \left(y^2\right)^2 - \left(\frac{3}{2}\right)^2 $

Step 3: Geometric Verification of the Identity (Visual Proof)

The algebraic identity $(a-b)(a+b) = a^2 - b^2$ can be geometrically proven by analyzing the area of a square of side $a$ with a smaller square of side $b$ removed. The remaining area can be rearranged into a rectangle with dimensions $(a+b)$ and $(a-b)$.

Step 4: Execution of the Algebraic Expansion & Simplification

We now simplify the expression $\left(y^2\right)^2 - \left(\frac{3}{2}\right)^2$ by applying the fundamental laws of exponents.

1. Simplifying the first term $\left(y^2\right)^2$:

According to the Power of a Power Property, $(x^m)^n = x^{m \cdot n}$. Therefore, we multiply the exponents:

$ \left(y^2\right)^2 = y^{2 \times 2} = y^4 $

2. Simplifying the second term $\left(\frac{3}{2}\right)^2$:

According to the Power of a Quotient Property, $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$. We distribute the square to both the numerator and the denominator:

$ \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4} $

Step 5: Final Polynomial Construction

By combining the simplified terms from Step 4 back into our expanded equation structure, we arrive at the final evaluated polynomial.

$ y^4 - \frac{9}{4} $

Final Solution: The product of the given binomials is $y^4 - \frac{9}{4}$.

Step 1: Algebraic Formulation and Base Selection

To evaluate the product $104 \times 96$ without performing direct arithmetic multiplication, we must express both factors in terms of a common, easily computable base. Observing the numerical values, both $104$ and $96$ are symmetrically distributed around the base value of $100$.

• The first factor can be written as an addition to the base: $104 = 100 + 4$
• The second factor can be written as a subtraction from the base: $96 = 100 - 4$

Substituting these expressions back into the original product, we obtain:

$104 \times 96 = (100 + 4)(100 - 4)$

Step 2: Application of the Difference of Squares Identity

The formulated expression $(100 + 4)(100 - 4)$ perfectly matches the structure of a fundamental polynomial identity [Per the algebraic identity for the product of the sum and difference of two terms].

The standard identity is defined as:

$(a + b)(a - b) = a^2 - b^2$

By mapping our specific variables to the identity, we establish:

• $a = 100$
• $b = 4$

Applying the identity to our expression yields:

$(100 + 4)(100 - 4) = (100)^2 - (4)^2$

Step 3: Geometric Visualization of the Identity

The algebraic identity $(a+b)(a-b) = a^2 - b^2$ can be rigorously proven through geometric area conservation. The area of a rectangle with dimensions $(a+b)$ and $(a-b)$ is mathematically equivalent to the area of a large square of side $a$ minus the area of a smaller square of side $b$.

Step 4: Computation of Squares and Final Evaluation

We now compute the numerical values of the squared terms derived in Step 2:

• Square of the base term: $(100)^2 = 100 \times 100 = 10000$
• Square of the deviation term: $(4)^2 = 4 \times 4 = 16$

Substitute these calculated values back into the difference equation:

$10000 - 16$

Performing the final subtraction operation:

$10000 - 16 = 9984$

Final Solution: 9984

Given Variables & Theoretical Foundation

We are given the volume of a cuboid expressed as a polynomial in terms of a variable $x$:

$V(x) = 3x^2 - 12x$

[Per the geometric definition of a cuboid], the volume $V$ is the product of its three mutually perpendicular spatial dimensions: Length ($L$), Width ($W$), and Height ($H$). Mathematically, this is expressed as:

$V = L \times W \times H$

To find the possible expressions for the dimensions of the cuboid, we must factorize the given binomial polynomial into three distinct linear or constant factors.

Step 1: Identifying the Greatest Common Factor (GCF)

We begin by analyzing the terms of the polynomial $3x^2 - 12x$ to extract the Greatest Common Factor (GCF). We decompose each term into its prime numerical and algebraic factors.

By comparing the factorizations, we identify the common elements:

• Numerical GCF: The greatest integer that divides both $3$ and $-12$ is $3$.
• Algebraic GCF: The highest power of $x$ common to both $x^2$ and $x$ is $x^1$ (or simply $x$).

Therefore, the overall Greatest Common Factor is $3x$.

Step 2: Algebraic Factorization

[By the Distributive Property of Multiplication over Addition], we can factor out the GCF from the original expression:

$V(x) = 3x^2 - 12x$

$V(x) = 3x(x) - 3x(4)$

$V(x) = 3x(x - 4)$

We have now successfully expressed the binomial as a product of its irreducible factors.

Step 3: Mapping Factors to Geometric Dimensions

The factored form of the volume is $3 \cdot x \cdot (x - 4)$. Because the volume of a cuboid requires three dimensions ($L \times W \times H$), we can directly map these three distinct factors to the dimensions of the cuboid.

• Dimension 1: $3$
• Dimension 2: $x$
• Dimension 3: $(x - 4)$

Note: Because multiplication is commutative ($A \times B \times C = B \times C \times A$), any of these expressions can represent the length, width, or height.

Geometric Visualization

Below is a high-precision spatial representation of the cuboid with its corresponding dimensional expressions.

Final Solution: The possible expressions for the dimensions of the cuboid are $3$, $x$, and $(x - 4)$.

Given Expression & Algebraic Foundation

We are tasked with expanding the following binomial cube:

$(2x + 1)^3$

To expand this expression, we utilize the standard algebraic identity for the cube of a binomial sum. [Per the fundamental theorem of polynomial expansion], the identity is defined as:

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

Step 1: Variable Identification and Substitution

By comparing the given expression $(2x + 1)^3$ with the standard identity $(a + b)^3$, we can establish a direct mapping of the terms:

• Let $a = 2x$
• Let $b = 1$

Substituting these values into the expansion formula yields:

$(2x + 1)^3 = (2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + (1)^3$

Step 2: Term-by-Term Simplification

We now systematically simplify each of the four terms generated by the expansion. [Applying the laws of exponents, specifically the power of a product rule: $(xy)^n = x^n y^n$].

• First Term ($a^3$):
  $(2x)^3 = 2^3 \cdot x^3 = 8x^3$


• Second Term ($3a^2b$):
  $3(2x)^2(1) = 3 \cdot (2^2 \cdot x^2) \cdot 1 = 3 \cdot 4x^2 \cdot 1 = 12x^2$


• Third Term ($3ab^2$):
  $3(2x)(1)^2 = 3 \cdot 2x \cdot 1 = 6x$


• Fourth Term ($b^3$):
  $(1)^3 = 1 \cdot 1 \cdot 1 = 1$

Step 3: Final Assembly of the Polynomial

Having simplified each individual component, we combine them sequentially to form the final expanded polynomial in standard form (descending order of degree):

$8x^3 + 12x^2 + 6x + 1$

Final Solution: The expanded form of $(2x + 1)^3$ is $8x^3 + 12x^2 + 6x + 1$.

Initial Setup & Theoretical Foundation

The geometric area of a rectangle is defined by the product of its two adjacent spatial dimensions, length and breadth. Mathematically, this is expressed as:

[$Area = Length \times Breadth$]

We are given the area of a rectangle as a quadratic polynomial in terms of the variable $y$:

$Area = 35y^2 + 13y - 12$

To find the possible expressions for the length and breadth, we must factorize this quadratic polynomial into the product of two linear binomials. Each resulting binomial will represent one of the dimensions of the rectangle.

Step 1: Coefficient Analysis and the Product-Sum Method

We will factorize the quadratic polynomial $ay^2 + by + c$ using the method of splitting the middle term. First, we identify the coefficients:

• Leading coefficient ($a$) = $35$
• Middle coefficient ($b$) = $13$
• Constant term ($c$) = $-12$

According to the product-sum factorization theorem, we must find two real numbers, let's call them $p$ and $q$, such that:

1. Their product equals $a \times c$:
$p \times q = 35 \times (-12) = -420$

2. Their sum equals $b$:
$p + q = 13$

Step 2: Prime Factorization to Determine Split Values

To systematically find the values of $p$ and $q$, we analyze the prime factorization of the absolute value of the product ($420$):

$420 = 2 \times 210 = 2^2 \times 105 = 2^2 \times 3 \times 35 = 2^2 \times 3 \times 5 \times 7$

We need to group these prime factors into two numbers whose difference is $13$ (since the product is negative, one number must be positive and the other negative). Let us test combinations:

• Combination 1: $(2^2 \times 5) = 20$ and $(3 \times 7) = 21$. Difference is $1$. (Incorrect)
• Combination 2: $(2^2 \times 7) = 28$ and $(3 \times 5) = 15$. Difference is $13$. (Correct)

Since the sum must be positive ($+13$), the larger number must be positive. Therefore, our two numbers are $28$ and $-15$.

[$28 \times (-15) = -420$ and $28 + (-15) = 13$]

Step 3: Splitting the Middle Term and Grouping

We substitute the middle term $13y$ with $28y - 15y$ in the original polynomial:

$35y^2 + 28y - 15y - 12$

Next, we apply factorization by grouping. We group the first two terms and the last two terms:

$(35y^2 + 28y) - (15y + 12)$

Extract the greatest common divisor (GCD) from each group:

For $(35y^2 + 28y)$, the GCD is $7y$:
$7y(5y + 4)$

For $-(15y + 12)$, the GCD is $-3$:
$-3(5y + 4)$

Now, substitute these back into the expression:

$7y(5y + 4) - 3(5y + 4)$

Notice that the binomial $(5y + 4)$ is a common factor. Factoring it out yields the final product of two linear expressions:

$(7y - 3)(5y + 4)$

Step 4: Geometric Representation and Dimensional Assignment

Because multiplication is commutative [by the Commutative Property of Multiplication, $A \times B = B \times A$], either of these binomial factors can represent the length, and the other will represent the breadth.

Final Solution: The possible expressions for the dimensions of the rectangle are Length = $(7y - 3)$ and Breadth = $(5y + 4)$, or vice versa.

Initial Setup & Algebraic Transformation

We are tasked with evaluating the numerical product $103 \times 107$ without executing direct multiplication. To achieve this, we must transform the arithmetic operation into an algebraic framework by expressing each factor as a binomial with a common base.

We decompose the given numbers using the base $100$:

• $103 = 100 + 3$
• $107 = 100 + 7$

Thus, the product can be rewritten as:

$103 \times 107 = (100 + 3)(100 + 7)$

Step 1: Selection of the Algebraic Identity

The expression $(100 + 3)(100 + 7)$ perfectly matches the standard polynomial identity for the product of two binomials sharing a common term. [Per the fundamental algebraic identity of binomial expansion]:

$(x + a)(x + b) = x^2 + (a + b)x + ab$

Step 2: Geometric Visualization of the Identity

To rigorously prove the spatial validity of this identity, we can represent the product as the area of a rectangle with dimensions $(x + a)$ and $(x + b)$. The total area is the sum of four distinct sub-rectangles.

Step 3: Variable Assignment and Substitution

By mapping our specific numerical values to the variables in the identity, we establish the following parameters:

• $x = 100$
• $a = 3$
• $b = 7$

Substituting these values into the identity $(x + a)(x + b) = x^2 + (a + b)x + ab$, we obtain:

$(100 + 3)(100 + 7) = (100)^2 + (3 + 7)(100) + (3 \times 7)$

Step 4: Algebraic Expansion and Arithmetic Computation

We now evaluate each term of the expanded polynomial sequentially [adhering to the standard order of operations, PEMDAS/BODMAS]:

• First Term ($x^2$): $(100)^2 = 100 \times 100 = 10000$
• Middle Term ($(a+b)x$): $(3 + 7) \times 100 = 10 \times 100 = 1000$
• Last Term ($ab$): $3 \times 7 = 21$

Step 5: Final Summation

We synthesize the evaluated terms to find the total product:

$10000 + 1000 + 21$

$= 11000 + 21$

$= 11021$

Final Solution: The evaluated product of $103 \times 107$, derived using algebraic identities without direct multiplication, is $11021$.

Step 1: Algebraic Setup and Identity Identification

We are tasked with factorising the following multivariable polynomial:

$27x^3 + y^3 + z^3 - 9xyz$

By observing the structure of the expression, we can see it consists of a sum of three perfect cubes followed by a product term. This specific structure corresponds directly to the standard algebraic identity for the sum of three cubes:

[Standard Algebraic Identity]
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Step 2: Term-by-Term Structural Mapping

To apply the identity, we must rewrite the given polynomial so that each term perfectly matches the $a^3$, $b^3$, $c^3$, and $-3abc$ components of the formula. We extract the cube roots of the first three terms:

• For the first term: $27x^3 = (3^3)(x^3) = (3x)^3$ [By the power of a product property: $x^n y^n = (xy)^n$]
• For the second term: $y^3 = (y)^3$
• For the third term: $z^3 = (z)^3$

Now, we verify if the fourth term, $-9xyz$, matches the $-3abc$ structure using our derived values for $a$, $b$, and $c$:

$-3(3x)(y)(z) = -9xyz$

Since the product matches the original expression exactly, we can definitively establish our variable mappings:

Step 3: Substitution into the Algebraic Identity

We now substitute $a = 3x$, $b = y$, and $c = z$ into the right-hand side of our identity:

$(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Substituting the mapped variables yields:

$= (3x + y + z) \left[ (3x)^2 + (y)^2 + (z)^2 - (3x)(y) - (y)(z) - (z)(3x) \right]$

Step 4: Simplification of the Polynomial Factors

To finalize the factorisation, we must expand and simplify the terms within the second bracket:

• Square the first term: $(3x)^2 = 9x^2$
• Square the second term: $(y)^2 = y^2$
• Square the third term: $(z)^2 = z^2$
• Compute the first cross-product: $-(3x)(y) = -3xy$
• Compute the second cross-product: $-(y)(z) = -yz$
• Compute the third cross-product: $-(z)(3x) = -3zx$

Reassembling the simplified terms into the second bracket, we get:

$(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$

Combining this with the first bracket gives us the complete, fully factorised expression.

Final Solution: The factorised form of the polynomial $27x^3 + y^3 + z^3 - 9xyz$ is $(3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.

Initial Setup & Theoretical Foundation

The area of a rectangle is geometrically defined as the product of its two adjacent spatial dimensions: length ($l$) and breadth ($b$). This relationship is expressed by the formula:

$\text{Area} = l \times b$

We are given the area of the rectangle in the form of a quadratic polynomial in terms of the variable $a$:

$\text{Area} = 25a^2 - 35a + 12$

[Per the Fundamental Theorem of Algebra and polynomial factorization principles], to find the possible expressions for the length and breadth, we must factorize this quadratic polynomial into the product of two linear binomials. These two resulting factors will represent the possible dimensions of the rectangle.

Step 1: Identifying the Factorization Method

We will factorize the quadratic polynomial $P(a) = 25a^2 - 35a + 12$ using the method of splitting the middle term. We compare the given polynomial to the standard quadratic form $Ax^2 + Bx + C$ to identify the coefficients:

• $A = 25$ (Coefficient of $a^2$)
• $B = -35$ (Coefficient of $a$)
• $C = 12$ (Constant term)

Step 2: Splitting the Middle Term

To split the middle term, we must find two real numbers, let us call them $p$ and $q$, that satisfy two specific conditions simultaneously:

1. Their product must equal $A \times C$:
   $p \times q = 25 \times 12 = 300$
2. Their sum must equal $B$:
   $p + q = -35$

Because the product ($300$) is positive and the sum ($-35$) is negative, [by the rules of integer arithmetic], both numbers $p$ and $q$ must be negative.

Let us evaluate the factor pairs of $300$ to find the correct combination:

The numbers $-15$ and $-20$ satisfy both conditions perfectly. Therefore, we will split the middle term $-35a$ into $-20a$ and $-15a$.

Step 3: Factorization by Grouping

We substitute the split terms back into the original polynomial:

$P(a) = 25a^2 - 20a - 15a + 12$

Next, we group the terms into pairs to extract the greatest common monomial from each group:

$P(a) = (25a^2 - 20a) - (15a - 12)$

Factor out the greatest common divisor (GCD) from the first group ($5a$) and the second group ($3$):

$P(a) = 5a(5a - 4) - 3(5a - 4)$

Notice that the binomial $(5a - 4)$ is now a common factor in both terms. We factor out $(5a - 4)$ [by the Distributive Property of Multiplication over Addition]:

$P(a) = (5a - 4)(5a - 3)$

Geometric Visualization

The factorization proves that a rectangle with an area of $25a^2 - 35a + 12$ can be constructed with sides measuring $(5a - 4)$ and $(5a - 3)$. Below is the geometric representation of this relationship.

Final Conclusion

Because multiplication is commutative ($l \times b = b \times l$), either of the two binomial factors can represent the length, and the remaining factor will represent the breadth.

Final Solution: The possible expressions for the length and breadth of the rectangle are $(5a - 3)$ and $(5a - 4)$.

Given Expression & Theoretical Foundation

We are tasked with expanding the binomial cube: $(2a - 3b)^3$.

To expand a binomial raised to the third power, we utilize the standard algebraic identity for the cube of a difference. [Per the fundamental theorems of polynomial expansion], the identity is defined as:

$(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$

Alternatively, it can be factored as $(x - y)^3 = x^3 - y^3 - 3xy(x - y)$. For standard polynomial expansion, the fully distributed form is preferred.

Step 1: Identification and Mapping of Variables

By comparing our given expression $(2a - 3b)^3$ with the standard identity $(x - y)^3$, we map the variables as follows:

• $x = 2a$
• $y = 3b$

Step 2: Substitution into the Algebraic Identity

Substitute $x = 2a$ and $y = 3b$ into the expanded form of the identity:

$(2a - 3b)^3 = (2a)^3 - 3(2a)^2(3b) + 3(2a)(3b)^2 - (3b)^3$

Step 3: Expansion and Simplification of Individual Terms

We will now apply the laws of exponents [specifically $(uv)^n = u^n v^n$] to evaluate each term systematically:

• First Term ($x^3$):
  $(2a)^3 = 2^3 \cdot a^3 = 8a^3$


• Second Term ($-3x^2y$):
  $-3(2a)^2(3b) = -3(4a^2)(3b)$
  Multiply the numerical coefficients: $-3 \cdot 4 \cdot 3 = -36$
  Multiply the variables: $a^2 \cdot b = a^2b$
  Result: $-36a^2b$


• Third Term ($+3xy^2$):
  $3(2a)(3b)^2 = 3(2a)(9b^2)$
  Multiply the numerical coefficients: $3 \cdot 2 \cdot 9 = 54$
  Multiply the variables: $a \cdot b^2 = ab^2$
  Result: $+54ab^2$


• Fourth Term ($-y^3$):
  $-(3b)^3 = -(3^3 \cdot b^3) = -27b^3$

Step 4: Final Assembly of the Polynomial

Combine all the simplified terms from Step 3 in descending order of the degree of $a$:

$(2a - 3b)^3 = 8a^3 - 36a^2b + 54ab^2 - 27b^3$

Final Solution: The expanded form of the cube $(2a - 3b)^3$ is $8a^3 - 36a^2b + 54ab^2 - 27b^3$.

Initial Setup & Algebraic Objective

We are tasked with verifying the following fundamental algebraic identity:

$x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$

To prove this rigorously, we will operate on the Right-Hand Side (RHS) of the equation and demonstrate, through sequential expansion and simplification, that it is mathematically equivalent to the Left-Hand Side (LHS).

Step 1: Expansion of the Squared Binomials on the RHS

Consider the Right-Hand Side (RHS) of the given equation:

$\text{RHS} = \frac{1}{2}(x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$

We begin by expanding the three squared binomial terms inside the square brackets. [Per the standard algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$], we obtain:

• $(x - y)^2 = x^2 - 2xy + y^2$
• $(y - z)^2 = y^2 - 2yz + z^2$
• $(z - x)^2 = z^2 - 2zx + x^2$

Step 2: Aggregation and Simplification of the Expanded Terms

Substitute these expanded forms back into the bracketed expression:

$[(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)]$

Next, we group the like terms (the squared variables):

$= (x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) - 2xy - 2yz - 2zx$

$= 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx$

By factoring out the greatest common scalar multiplier ($2$), the expression simplifies to:

$= 2(x^2 + y^2 + z^2 - xy - yz - zx)$

Step 3: Cancellation of the Scalar Multiplier

Now, substitute this simplified bracketed expression back into the full RHS equation:

$\text{RHS} = \frac{1}{2}(x + y + z) \cdot \left[ 2(x^2 + y^2 + z^2 - xy - yz - zx) \right]$

The scalar fraction $\frac{1}{2}$ and the factored integer $2$ cancel each other out exactly ($\frac{1}{2} \times 2 = 1$), yielding:

$\text{RHS} = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

Note: At this stage, the expression matches the standard factorization of $x^3 + y^3 + z^3 - 3xyz$. However, for absolute rigor, we will perform the full polynomial distribution to prove the equivalence to the LHS.

Step 4: Final Polynomial Distribution

We distribute the trinomial $(x + y + z)$ across the polynomial $(x^2 + y^2 + z^2 - xy - yz - zx)$ by multiplying each term in the first bracket by every term in the second bracket:

Distributing $x$:
$x(x^2 + y^2 + z^2 - xy - yz - zx) = x^3 + xy^2 + xz^2 - x^2y - xyz - zx^2$

Distributing $y$:
$y(x^2 + y^2 + z^2 - xy - yz - zx) = x^2y + y^3 + yz^2 - xy^2 - y^2z - xyz$

Distributing $z$:
$z(x^2 + y^2 + z^2 - xy - yz - zx) = zx^2 + zy^2 + z^3 - xyz - yz^2 - xz^2$

Now, we sum all the distributed terms together:

$= x^3 + y^3 + z^3$
$\quad + xy^2 - xy^2$
$\quad + xz^2 - xz^2$
$\quad - x^2y + x^2y$
$\quad - zx^2 + zx^2$
$\quad + yz^2 - yz^2$
$\quad - y^2z + zy^2$
$\quad - xyz - xyz - xyz$

Observe the systematic cancellation of the intermediate cross-terms:

• $xy^2$ cancels with $-xy^2$
• $xz^2$ cancels with $-xz^2$
• $-x^2y$ cancels with $x^2y$
• $-zx^2$ cancels with $zx^2$
• $yz^2$ cancels with $-yz^2$
• $-y^2z$ cancels with $zy^2$ (since $zy^2 = y^2z$)

The only terms that survive the cancellation are the cubic terms and the three $-xyz$ terms:

$= x^3 + y^3 + z^3 - 3xyz$

This resulting expression is exactly the Left-Hand Side (LHS) of our initial equation.

Final Solution: By expanding the right-hand side and systematically canceling the intermediate polynomial terms, we have rigorously proven that $\text{RHS} = \text{LHS}$. Therefore, the identity $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$ is verified.