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

(N/A) Given that $x+y+z=0.$
Therefore,$x+y=-z.$
Cubing both sides,we get $(x+y)^{3}=(-z)^{3}.$
Using the identity $(x+y)^{3} = x^{3}+y^{3}+3xy(x+y),$ we have:
$x^{3}+y^{3}+3xy(x+y) = -z^{3}.$
Since $x+y=-z,$ substitute this into the equation:
$x^{3}+y^{3}+3xy(-z) = -z^{3}.$
$x^{3}+y^{3}-3xyz = -z^{3}.$
Rearranging the terms,we get:
$x^{3}+y^{3}+z^{3} = 3xyz.$
Hence,if $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}=3xyz.$