If \(x,y,z\) are different and \(\triangle=\left|\begin{array}{lll}x&x^2&1+x^3\\ y&y^2&1+y^3\\ z& z^2 &1+z^3\end{array}\right|=0\) then

\(1+xyz=0\)

\(\triangle=\left|\begin{array}{lll}x&x^2&1+x^3\\ y&y^2&1+y^3\\ z& z^2 &1+z^3\end{array}\right|=0\)

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

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

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

so $1+xyz=0$