Question

If $x+y+z=0$ then $x^3+y^3+z^3=?$

Answer 1

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

$x+y+z=0$

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

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

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