Question

Verify that $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z)+(z-x{)}^2]$

Answer 1

We have, $L.H.S.=x^3+y^3+z^3-3xyz$

$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ [by poynomial identity]

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

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

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

Hence proved.