Question

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

Answer 1

Prove the expression

It is given that $x+y+z=0$

We know that,

$$\begin{gathered} x^3+y^3+z^3–3xyz=(x+y+z)(x^2+y^2+z^2–xy–yz–xz) \\ \Rightarrow x^3+y^3+z^3–3xyz=\left(0\right)(x^2+y^2+z^2–xy–yz–xz)\left[\because x+y+z=0\right] \\ \Rightarrow x^3+y^3+z^3–3xyz=0 \\ \Rightarrow x^3+y^3+z^3=3xyz \end{gathered}$$

Hence, $x^3+y^3+z^3=3xyz$.