Given that $xyz=-1, $ the value of the determinant $\begin{bmatrix}x & x^2 & 1+x^3\\y & y^2 & 1+y^3\\z & z^2 & 1+z^2\end{bmatrix}$, is

0

The correct answer is option (1) : 0

We have,

$\begin{bmatrix}x & x^2 & x^3\\y & y^2 & 1+y^3\\z & z^2 & 1+z^3\end{bmatrix}$

$=\begin{bmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{bmatrix}+\begin{bmatrix}x & x^2 & +x^3\\y & y^2 & y^3\\z & z^2 & z^3\end{bmatrix}$         $\begin{matrix} \text{Since each element of third}\\\text{column is sum of two elements}\end{matrix}$

$=\begin{bmatrix}x & x^2 & 1\\y & y^2 & 1\\z & z^2 & 1\end{bmatrix}+xyz\begin{bmatrix}1 & x & +x^2\\1 & y & y^2\\1 & z & z^2\end{bmatrix}$         $\begin{matrix} \text{Taking x, y and z common}\\C_1, C_2 \, and \, C_3\, \text{in second det}\end{matrix}$

$=\begin{bmatrix}x & 1 & x^2\\y & 1 & y^2\\z & 1 & z^2\end{bmatrix}+xyz\begin{bmatrix}1 & x & +x^2\\1 & y & y^2\\1 & z & z^2\end{bmatrix}$   $\begin{matrix} Interchanging \, C_2 \\ and \, C_3 \, in\, first \, det \end{matrix}$

$=\begin{bmatrix}1 & x & x^2\\1 & y & y^2\\1 & z & z^2\end{bmatrix}+xyz\begin{bmatrix}1 & x & +x^2\\1 & y & y^2\\1 & z & z^2\end{bmatrix}$   $\begin{matrix} Interchanging \, C_1 \\ and \, C_2 \, in\, first \, det \end{matrix}$

$=\begin{bmatrix}1 & x & x^2\\1 & y & y^2\\1 & z & z^2\end{bmatrix}(1+xyz)= 0 $    $[∵xyz=-1]$