Question

Verify that -
${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \frac{1}{2} (x + y + z)( {( x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} )$
REFERENCE -
Class 9 - Exercise 2.5 - Question 12
(Explanation - Mandatory)

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Answer 1

Step-by-step explanation:

LHS = x³ + y³ + z³ - 3xyz

We know that,

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)

RHS = (1/2)(x + y + z)[(x - y)² + (y - z)² + (z - x)²]

= (1/2)(x + y + z)[(x² - 2xy + y²) + (y² - 2yz + z²) + (z² - 2xz + x²)]

= (1/2)(x + y + z)[x² - 2xy + y² + y² - 2yz + z² + z² - 2xz + x²]

= (1/2)(x + y + z)[x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2xz]

= (1/2)(x + y + z)[2x² + 2y² + 2z² - 2xy - 2yz - 2xz]

= (1/2)(x + y + z)(2)[x² + y² + z² - xy - yz - xz]

= (1/2)(2)(x + y + z)[x² + y² + z² - xy - yz - xz]

= (x + y + z)[x² + y² + z² - xy - yz - xz]

LHS = RHS

Hence Verified.

Hope it helped and you understood it........All the best

Answer 2

GIVEN :-

• x³ + y³ + z³ - 3xyz = 1/2 [(x + y + z){(x - y)² + (y - z)² + (z - x)²

TO VERIFY :-

• L.H.S = R.H.S.

SOLUTION :-

L.H.S,

$\\ : \implies \displaystyle \sf \: x ^{3} + y ^{3} + z ^{3} - 3xyz$

$\bigstar \: \displaystyle \sf identity \to\: x ^{3} + y ^{3} + z ^{3} - 3xyz = (x + y + z)(x ^{2} + y ^{2} + z ^{2} - xy - yz - zx)$

$: \implies \displaystyle \sf \bigg(x + y + z \bigg) \bigg(x ^{2} + y ^{2} + z ^{2} - xy - yz - zx \bigg)$

$: \implies \displaystyle \sf \dfrac{1}{2} \bigg(x + y + z \bigg)\Bigg[2x ^{2} + 2y ^{2} + 2z ^{2} -2 xy - 2yz - 2zx \Bigg]$

$:\implies \displaystyle \sf \dfrac{1}{2} \bigg(x + y + z \bigg)\Bigg[x ^{2} + x ^{2} + y ^{2} + y ^{2} + z ^{2} + z ^{2} -2 xy - 2yz - 2zx \Bigg]$

On manipulating the terms,

$\\ : \implies \displaystyle \sf \dfrac{1}{2} \bigg(x + y + z \bigg)\Bigg[ \bigg(x ^{2} + y ^{2} - 2xy \bigg) +\bigg(y ^{2} + z ^{2} - 2yz \bigg) +\bigg(z ^{2} - x ^{2} - 2xz \bigg)\Bigg]$

$: \implies \displaystyle \sf \dfrac{1}{2} \bigg(x + y + z \bigg)\Bigg[ \bigg(x - y \bigg) ^{2} + \bigg(y - z \bigg) ^{2} + \bigg(z - x \bigg) ^{2} \Bigg]$

Here we got,

• L.H.S = R.H.S

$\\ \\ \boxed{ \boxed{ \boxed{\displaystyle \large{\mathbb{ \red{HENCE \: \: VERIFIED}}}}}}$