Aspire Faculty ID #16707 · Topic: CUET 2023 · Just now
CUET 2023

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.

Solution

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, $xyz+1=0$ $\Rightarrow xyz=-1$

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