Question

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

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

Answer 1

Answer: (B)

$\begin{vmatrix}x^{3}+1&x^{2}&x\\ y^{3}+1&y^{2}&y\\ z^{3}+1&z^{2}&z\end{vmatrix} = 0 $
$\Rightarrow \, \begin{vmatrix}x^{3}&x^{2}&x\\ y^{3}&y^{2}&y\\ z^{3}&z^{2}&z\end{vmatrix} + \begin{vmatrix}1&x^{2}&x\\ 1&y^{2}&y\\ 1&z^{2}&z\end{vmatrix} = 0$
$ \Rightarrow xyz \begin{vmatrix}x^{2}&x&1\\ y^{2}&y&1\\ z^{2}&z&1\end{vmatrix} + \begin{vmatrix}x^{2}&x&1\\ y^{2}&y&1\\ z^{2}&z&1\end{vmatrix}=0 $
$ \Rightarrow xyz = - 1$
$ \left(\because \begin{vmatrix}x^{2}&x&1\\ y^{2}&y&1\\ z^{2}&z&1\end{vmatrix} = -\left(x-y\right)\left(y-z\right)\left(z-x\right)\ne0 \right)$