Question

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

Answer 1

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

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

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

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

$=(x+y+z)[x^2+y^2+z^2-xy-yz-xz]$

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

Hence proved.