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Q12:
Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
Solution :
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.
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 : $27x^3 + y^3 + z^3 – 9xyz$
- 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) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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