Question

If $x, y$ and $z$ are all distinct and $\begin{vmatrix}x&x^{2}&1+x^{3}\\ y&y^{2}&1+y^{3}\\ z&z^{2}&1+z^{3}\end{vmatrix} = 0 $ then the value of $xyz$ is

• A. -2
• B. -1
• C. -3
• D. None of these

Answer 1

Answer: (B)

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