Question

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

• A. $-1$
• B. $-2$
• C. $2$
• D. $1$

Answer 1

The correct option is A $-1$

Since each element of $C_1$ is the sum of two elements, putting the determinant as sum of two determinants,
we get,

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

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

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

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

Since $\Delta =0$ and $x,y,z$ all are distinct, we have $xyz+1=0\text{or}xyz=-1$