If (x + y + z) = 0, then prove that (x^{3} + y^{3} + z^{3}) = 3xyz.

x+y+z = 0

⇒x+y = -z

(x+y)^{3}= (-z)^{3}

x^{3}+y^{3}+3xy(x+y)= -z^{3}

x^{3}+y^{3}+3xy(-z)= -z^{3} [x+y= -z]

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

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

Hence, x+y+z= 0

⇒x^{3}+y^{3}+z^{3}= 3xyz