Question

If $x\ne y\ne z$ 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 $xyz$ will be

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

Answer: (C)

$-1$

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

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

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

$\left(xy\right)\left(yz\right)\left(zx\right)(1+xyz)=0$

but $x\ne y\ne z$ so $xyz=-1$