# If x + y + z = 9 and xy + yz + zx = 23,

Question:

If $x+y+z=9$ and $x y+y z+z x=23$, then the value of $\left(x^{3}+y^{3}+z^{3}-3 x y z\right)=?$

(a) 108
(b) 207
(c) 669
(d) 729

Solution:

(a) 108

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

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

$=9 \times(81-3 \times 23)$

$=9 \times 12$

$=108$