Question

If $x,y,z$ are different and $\left|\begin{array}{ccc}x& x^2& 1+x^2\\ y& y^2& 1+y^2\\ z& z^2& 1+z^2\end{array}\right|=0$ then prove that $1+xyz=0$.

Answer 1

$\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& {1+z}^3\end{array}\right|\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0C_1\leftrightarrow C_3-\left|\begin{array}{ccc}1& x^2& x\\ 1& y^2& y\\ 1& z^2& z\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0C_2\leftrightarrow C_3\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=01+xyz=0$