Question

verify that x cube + y cube + z cube -3xyz = 1/2 (x+y+z) [(x-y)square + (y-z) square + (z-x)square ]

Answer 1

here is your answer buddy, hope it helps you

Answer 2

Answer:

We have to verify:

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

Consider,

LHS = x³ + y³ + z³ - 3xyz

RHS

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

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

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

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

$=\frac{1}{2}(x+y+z)\times2\times(x^2+y^2+z^2-xy-yx-zx)$

$=\frac{1}{2}\times2(x+y+z)(x^2+y^2+z^2-xy-yx-zx)$

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

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

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

LHS = RHS

Hence Verified.