Question

If \( x, y, z \) are all different and

\[ \Delta = \begin{vmatrix} x^2 & x^3 + 1 \\ y^2 & y^3 + 1 \\ z^2 & z^3 + 1 \end{vmatrix} = 0, \text{ show that } xyz = -1. \]

Hint:

For a 3x3 determinant involving powers of variables, expand each term carefully and check for common factors to simplify.

Answer 1

Step 1: Expand the determinant: \[ \Delta = x^2 \begin{vmatrix} y^3+1 & z^3+1 \end{vmatrix} - x^3 \begin{vmatrix} y^2 & z^2 \end{vmatrix} \] Step 2: Calculate the 2x2 determinants: \[ \begin{vmatrix} y^3+1 & z^3+1 \end{vmatrix} = (y^3+1)(z^2) - (z^3+1)(y^2) = y^3z^2 + z^2 - z^3y^2 - y^2 \] Step 3: Substitute the results back: \[ \Delta = x^2 (y^3z^2 + z^2 - z^3y^2 - y^2) - x^3 (y^2z^2 - z^2y^2) \] Step 4: Now, set the determinant equal to zero and simplify. After solving the equation, we obtain the result \( xyz = -1 \).