Question

If $x,yandz$ are all 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:

• A. $xyz=-1$
• B. $xyz=1$
• C. $xyz=-2$
• D. $xyz=2$

Answer 1

The correct option is A

$xyz=-1$

Explanation for the correct option.

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

By using the property of determinant, we get

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

Hence, option A is correct.