Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $\pm 1$

Answer 1

The correct option is A $-1$

$\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow (1+xyz)\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$\Rightarrow (1+xyz)\left|\begin{array}{ccc}0& x-y& x^2-y^2\\ 0& y-z& y^2-z^2\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow (1+xyz)(x-y)(y-z)\left|\begin{array}{ccc}0& 1& x+y\\ 0& 1& y+z\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow (1+xyz)(x-y)(y-z)(z-x)=0$
$\Rightarrow xyz=-1$ ($\because x\ne y\ne z$)