If \(x, y, z\) are different and \(\Delta = \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix} = 0\), then the value of \(xyz\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −1
The third column is a sum, so the determinant splits into two determinants:
\( \displaystyle \Delta = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} \)
In the second determinant, take x, y and z out of rows 1, 2 and 3:
\( \displaystyle \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = xyz\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
In the first determinant, interchange C3 and C2, then C2 and C1. Two interchanges leave the sign unchanged:
\( \displaystyle \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
So
\( \displaystyle \Delta = (1 + xyz)\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
Applying R2 → R2 − R1 and R3 → R3 − R1:
\( \displaystyle \begin{vmatrix} 1 & x & x^2 \\ 0 & y - x & y^2 - x^2 \\ 0 & z - x & z^2 - x^2 \end{vmatrix} = (y - x)(z - x)\begin{vmatrix} 1 & y + x \\ 1 & z + x \end{vmatrix} \)
\(= (y - x)(z - x)(z - y)\)
\(= (x - y)(y - z)(z - x)\)
Hence
\(\Delta = (1 + xyz)(x - y)(y - z)(z - x) = 0\)
Since x, y, z are different, none of the factors \(x - y\), \(y - z\), \(z - x\) is zero. So
\(1 + xyz = 0\)
\(xyz = -1\)