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

Given: If $x+y+z = 0$

To find: Show that $x^{3}+y^{3}+z^{3}=3xyz$

Solution:

We know that,

 =$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

Putting $x+y+z =0$

$x^{3}+y^{3}+z^{3}-3xyz= (0)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$

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

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