Question

If x, y. z are distinct real numbers and $\left|\begin{array}{ccc}{x}^3+1& x^2& x\\ y^3+1& y^2& y\\ z^3+1& z^2& z\end{array}\right|=0$ then xyz is equal to

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

Answer 1

Answer: (B)

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