If $x+y+z=0$, show that $x^{3}+y^{3}+z^{3}=3 x y z$.

Solution:

It is known that,

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

Put $x+y+z=0$,

$x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

$x^{3}+y^{3}+z^{3}-3 x y z=0$

$x^{3}+y^{3}+z^{3}=3 x y z$