If $x,y,z$ 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 $xyz$ is.
🎥 Video solution / Text Solution of this question is given below:
Expand the determinant by using column operation:
$C_3 \rightarrow C_3 - C_1^3$
Then the determinant becomes a Vandermonde determinant multiplied by $(xyz+1)$.
Since $x,y,z$ are distinct, the Vandermonde determinant is non-zero.
Hence,
$xyz+1=0$
$\Rightarrow xyz=-1$