Question

If $x,y,z$ are different and $\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$ then prove that $xyz=-1$

Answer 1

Given:- $\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$

To prove:- $xyz=-1$

Proof:-

$=\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$

$=\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$

$=\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|=0$

$1+xyz=0$

$\Rightarrow xyz=-1$

Hence proved.