Question

If $x=2,y=1$ and $z=-3$, then $x^3+y^3+z^3-3xyz$ is equal to

• A. $6$
• B. $0$
• C. $2$
• D. $8$

Answer 1

The correct option is B $0$

Given,
$x+y+z$ $=2+1-3$$=0$
Thus,
$x+y+z=0$
$\Rightarrow x+y=-z$ ...(i)
$\Rightarrow (x+y{)}^3=(-z{)}^3$
$\Rightarrow x^3+y^3+3xy(x+y)=-z^3$
$\Rightarrow x^3+y^3+z^3+3xy(-z)=0$ from(i)
$\Rightarrow x^3+y^3+z^3-3xyz=0$

Hence, option B.