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

0

We have,

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

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

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

$=-\begin{vmatrix}x&1&x^2 \\y &1&y^2 \\z&1&z^2 \end{vmatrix}+xyz\begin{vmatrix}1&x &x^2\\1 &y & y^2\\1&z &z^2\end{vmatrix}$  [Interchanging $C_2$ and $C_3$ in first det]

$=\begin{vmatrix}1&x &x^2\\1 &y & y^2\\1&z &z^2\end{vmatrix}+xyz\begin{vmatrix}1&x &x^2\\1 &y & y^2\\1&z &z^2\end{vmatrix}$  [Interchanging $C_1$ and $C_2$ in first det]

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