Question

If $ x, y $ and $ z $ are all different 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

• A. $ xyz = -1 $
• B. $ xyz = 1 $
• C. $ xyz = -2 $
• D. $ xyz = 2 $

Answer 1

Answer: (A)

Given , $\left|\begin{matrix}x&x^{2}&1+x^{3}\\ y&y^{2}&1+y^{3}\\ z&z^{2}&1+z^{3}\end{matrix}\right|=0$
$\Rightarrow \left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|+\left|\begin{matrix}x&x^{2}&x^{3}\\ y&y^{2}&y^{3}\\ z&z^{2}&z^{3}\end{matrix}\right|=0 $
$\Rightarrow \left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|+xyz \left|\begin{matrix}1&x&x^{2}\\ 1&y&y^{2}\\ 1&z&z^{2}\end{matrix}\right|=0$
$\Rightarrow \left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|+xyz\left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|=0$
$\Rightarrow \left(1+xyz\right) \left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|=0$
$\Rightarrow \left(1+xyz\right)\left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|=0 $
$\Rightarrow 1+xyz=0$
and $ \left|\begin{matrix}x&x^{2}&1\\ y&y^{2}&1\\ z&z^{2}&1\end{matrix}\right|\ne0$
$\therefore xyz=-1$