Question

If $x+y+z=7,xy+yz+zx=16$, find the value of $x^3+y^3+z^3-3xyz$

Answer 1

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

$=(x+y+z)[x^2+y^2+z^2-(xy+yz+zx\left)\right]$

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

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

$=7[7^2-3(16\left)\right]$

$=7[49-48]$

$=7$