Question

$verify \: that \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \frac{1}{2} (x + y + z)( {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} )$

Solve it with all the steps.

Answer 1

LHS:

Given : x^3 + y^3 + z^3 - 3xyz.

We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

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

On Multiplying by '2', we get

$=> \frac{1}{2} * 2(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$

$=> \frac{1}{2}[(x + y + z)(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx)]$

$= > \frac{1}{2}(x + y + z)(x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx)$

$=> \frac{1}{2}(x + y + z)(x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx)$

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

Hope it helps!

Answer 2

$\boxed{HOLA!}$

$\bf{ANSWER \: IN \: THE \: ATTACHMENT}$