If x, y, z are all distinct and […] = 0 then the value of xyz is :

The correct answer is Option (2) → -1

$\begin{vmatrix} x & x^2 & 1+x^3\\y & y^2& 1+y^3\\z & z^2 & 1+z^3\end{vmatrix}=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+\begin{vmatrix} x & x^2 & x^3\\y & y^2& y^3\\z & z^2 & z^3\end{vmatrix}=0$

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

$(C_1↔C_3)$

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

$(C_1↔C_2)$

$=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+xyz\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}=0$

so $xyz=-1$