Q42 / 120 · VITEEE 2012 · Mathematics
If $x, y$ and $z$ are all distinct and $\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0$, then the value of $x y z$ is
- A$-2$
- Correct answerB$-1$
- C$-3$
- DNone of these
Solution
$\left|\begin{array}{ccc}\mathrm{x} & \mathrm{x}^{2} & 1+\mathrm{x}^{3} \\ \mathrm{y} & \mathrm{y}^{2} & 1+\mathrm{y}^{3} \\ \mathrm{z} & \mathrm{z}^{2} & 1+\mathrm{z}^{3}\end{array}\right|=0$
$$
\begin{array}{l}
\Rightarrow\left|\begin{array}{lll}
\mathrm{x} & \mathrm{x}^{2} & 1 \\
\mathrm{y} & \mathrm{y}^{2} & 1 \\
\mathrm{z} & \mathrm{z}^{2} & 1
\end{array}\right|+\left|\begin{array}{ccc}
\mathrm{x} & \mathrm{x}^{2} & \mathrm{x}^{3} \\
\mathrm{y} & \mathrm{y}^{2} & \mathrm{y}^{3} \\
\mathrm{z} & \mathrm{z}^{2} & \mathrm{z}^{3}
\end{array}\right|=0 \\
\Rightarrow\left|\begin{array}{lll}
\mathrm{x} & \mathrm{x}^{2} & 1 \\
\mathrm{y} & \mathrm{y}^{2} & 1 \\
\mathrm{z} & \mathrm{z}^{2} & 1
\end{array}\right|+\mathrm{xyz}\left|\begin{array}{ccc}
1 & \mathrm{x} & \mathrm{x}^{2} \\
1 & \mathrm{y} & \mathrm{y}^{2} \\
1 & \mathrm{z} & \mathrm{z}^{2}
\end{array}\right|=0
\end{array}
$$
$$
\Rightarrow(1+x y z)\left|\begin{array}{ccc}
x & x^{2} & 1 \\
y & y^{2} & 1 \\
z & z^{2} & 1
\end{array}\right|=0
$$
$$
\begin{array}{l}
\Rightarrow(1+x y z)\left[x\left(y^{2}-z^{2}\right)-y\left(x^{2}-z^{2}\right)\right. \\
\left.\quad+z\left(x^{2}-y^{2}\right)\right]=0 \\
\Rightarrow(1+x y z)(x-y)(y-z)(z-x)=0 \\
\Rightarrow 1+x y z=0 \Rightarrow x y z=-1
\end{array}
$$
Hence, the correct answer is option: 'B'.