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

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

Given that :

x + y + z = 0 …(1)

We know that,

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

Substituting the value of (1) in (2), we get

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

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