If x ≠ y ≠ z then […] is equal to

The correct answer is Option (3) → \((x-y)(y-z)(z - x)\)

Let \(\Delta\) be the determinant. Applying the row operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):

    \( \displaystyle \Delta = \begin{vmatrix}1&x&x^2\\0&y-x&y^2-x^2\\0&z-x&z^2-x^2\end{vmatrix} \)

Since \(y^2 - x^2 = (y - x)(y + x)\) and \(z^2 - x^2 = (z - x)(z + x)\), take out \((y - x)\) from \(R_2\) and \((z - x)\) from \(R_3\):

    \( \displaystyle \Delta = (y-x)(z-x)\begin{vmatrix}1&x&x^2\\0&1&y+x\\0&1&z+x\end{vmatrix} \)

Expanding along \(C_1\):

    \(\Delta = (y-x)(z-x)\left[(z + x) - (y + x)\right]\)

    \(= (y-x)(z-x)(z-y)\)

Since \((y - x)(z - y) = (x - y)(y - z)\):

    \(\Delta = (x-y)(y-z)(z-x)\)

Tip: This determinant, with rows \(1, x, x^2\); \(1, y, y^2\); \(1, z, z^2\), always equals \((x-y)(y-z)(z-x)\). It is a standard result worth remembering; it is 0 exactly when two of x, y, z are equal.