13.If show that x + y + z = 0, show that $$x^3 + y^3 + z^3 = 3xyz$$.
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$
by using identity $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
i.e., $$= (0)(x^2 + y^2 + z^2 - xy - yz - zx)$$
Therefore, $$x^3 + y^3 + z^3 = 3xyz$$