Question

If the determinant $\Delta =\left|\begin{array}{ccc}x& x^2& x^3-1\\ y& y^2& y^3-1\\ z& z^2& z^3-1\end{array}\right|$ is zero for distinct values of $x,y,z$, then the value of $4+xyz$ is

Answer 1

The given determinant can be written as,
$\Delta =\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|+\left|\begin{array}{ccc}x& x^2& -1\\ y& y^2& -1\\ z& z^2& -1\end{array}\right|={\Delta }_1+{\Delta }_2$ (say)
$\Rightarrow {\Delta }_1=xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$
(taking $x,y,z$ common from $R_1,R_2,R_3$)
and ${\Delta }_2=(-1)(-1)\left|\begin{array}{ccc}-1& x& x^2\\ -1& y& y^2\\ -1& z& z^2\end{array}\right|$
(interchanging $C_2\leftrightarrow C_3$ and then $C_1\leftrightarrow C_2$ )
$\Rightarrow {\Delta }_2=(-1)\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$
(taking $-1$ common from $C_1$)
$\Rightarrow \Delta =(xyz-1)\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$
Since $\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=(x-y)(y-z)(z-x)\ne 0$ for distinct values of $x,y,z,$
$\Rightarrow xyz-1=0$
So, $xyz+4=5$