Question

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

Answer 1

Given, $x+y+z=0$

We know that

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ we have

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Now, $x+y+z=0,$ then

$x^3+y^3+z^3-3xyz=\left(0\right)(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz=0$

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

Hence, this is the answer.