Question

If $x=(\beta -\gamma )(\alpha -\delta ),y=(\gamma -\alpha )(\beta -\delta ),z=(\alpha -\beta )(\gamma -\delta )$, then the value of $x^3+y^3+z^3-3xyz$ is

• A. $0$
• B. ${\alpha }^6+{\beta }^6+{\gamma }^6+{\delta }^6$
• C. ${\alpha }^6{\beta }^6{\gamma }^6{\delta }^6$
• D. $1$

Answer 1

The correct option is A $0$

On simplification, we get $x+y+z=0$
We have,
$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
Hence in this case,
$x^3+y^3+z^3-3xyz=0$