Question

If $x,y$ and $z$ are all distinct 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 the value of $xyz$ is

• A. -2
• B. -1
• C. -3
• D. None of these

Answer 1

Answer: (B)

$\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& x^3\\ z& z^2& x^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$\Rightarrow \left(1+xyz\right)\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|=0$
$\Rightarrow \left(1+xyz\right)\left[x\left(y^2-z^2\right)-y\left(x^2-z^2\right)+z\left(x^2-y^2\right)\right]=0$
$\Rightarrow \left(1+xyz\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)=0$
$\Rightarrow 1+xyz=0\Rightarrow xyz=-1$