If $x\ne y\ne z\text{ and }\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$ , then x y z is equal to :

Let $A=\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|$

By taking x, y, z common from the rows $R_1,R_2\text{ and }{R}_3$ respectively. So,

$A=xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$

Operate ${\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_1\text{ and }{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1$

$\Rightarrow A=xyz\left|\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right|$

Now take common y- x and z - x from the rows  ${\mathrm{R}}_2\text{ and }{\mathrm{R}}_3$ respectively. Thus

$\begin{array}{l}A=xyz(y-x)(z-x)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& y+x\\ 0& 1& z+x\end{array}\right|\\ =xyz(y-x)(z-x)(z-y)\\ =xyz(x-y)(y-z)(z-x)\end{array}$

Given $\left|\mathrm{A}\right|=0$      

So $xyz=0\because x\ne y\ne z$ (given)