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

Given:

$x+y+z = 0$

To do:

We have to show that $x^{3}+y^{3}+z^{3}=3xyz$

Solution:

We know that,

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

Therefore,

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

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

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

 Hence proved.