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Q9(ii):
Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
Solution :
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.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q8(v): Factorise each of the following: (v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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