Question
Evaluate $x ^3+ y ^3+ z ^3-3 xyz$ when $x =-2, y =-1$ and $z =3$.
✓
Answer
We have:
$x^3+y^3+z^3-3 x y z$
$=(-2)^3+(-1)^3+(3)^3-3 \times(-2) \times(-1) \times 3$
$=-8-1+27-18=-27+27=0$
$x^3+y^3+z^3-3 x y z$
$=(-2)^3+(-1)^3+(3)^3-3 \times(-2) \times(-1) \times 3$
$=-8-1+27-18=-27+27=0$
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