Question

If $x+y+z=6$ and $xy+yz+zx=12$ then show that $x^3+y^3+z^3=3xyz$

Answer 1

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

$=\left(6\right)\left\{\right(x+y+z{)}^2-2xy-2yz-2zx-xy-yz-zx\}$

$=\left(6\right)\left\{\right(x+y+z{)}^2-3xy-3yz-3zx\}$

$=\left(6\right)\left\{\right(x+y+z{)}^2-3(xy+yz+zx)\}$

$=\left(6\right)(6^2-3\times 12)$

$=\left(6\right)(36-36)$

$=0$

Therefore,

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