Question

If $x\ne y\ne z$ and $\left|\begin{array}{ccc}x& x^2& 1+x^3\\ y& y^2& 1+y^3\\ z& z^2& 1+z^3\end{array}\right|=0$, the value of $xyz$

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C $-1$

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

Since each element of $C_1$ is the sum of two elements, putting the determinant as sum of two determinants, we get
$\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$
$\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|+xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|=0$
$(1+xyz)\left|\begin{array}{ccc}x& x^2& 1\\ y& y^2& 1\\ z& z^2& 1\end{array}\right|=0$
$R_1\to R_1-R_2,{R_2\to R}_2-R_3$
$(1+xyz)\left|\begin{array}{ccc}x-y& x^2-y^2& 0\\ y-z& y^2-z^2& 0\\ z& z^2& 1\end{array}\right|=0$
$(1+xyz)(x-y)(y-z)\left|\begin{array}{ccc}1& x+y& 0\\ 1& y+z& 0\\ z& z^2& 1\end{array}\right|=0$
$(1+xyz)(x-y)(y-z)(z-x)=0$
$\Rightarrow xyz=-1$