Question

If $x,yandz$ 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. $-4$
• B. $-1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is B $-1$

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

$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$ ... [Using property of determinant]

$\Rightarrow \left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|=0$ ... [Using property of determinant]

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

$\Rightarrow (1+xyz)(x-y)(y-z)(z-x)=0$

$\Rightarrow 1+xyz=0\Rightarrow xyz=-1$

Hence, option B is correct.