If x, y, z are distinct non zero real numbers, prove that Δ = |x x² yz; y y² zx; z z² xy| = |y² y³…

Multiply $R_1, R_2, R_3$ by $x, y, z$ respectively and divide by $xyz$: $\Delta = \frac{1}{xyz}\begin{vmatrix} x^2 & x^3 & xyz \\ y^2 & y^3 & xyz \\ z^2 & z^3 & xyz \end{vmatrix} = \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix}$ (taking $xyz$ common from $C_3$). Taking the transpose, $\Delta = \begin{vmatrix} x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \\ 1 & 1 & 1 \end{vmatrix}$. Interchanging $C_1, C_2$ and then $C_2, C_3$ (two interchanges, so the sign is unchanged) gives $\begin{vmatrix} y^2 & z^2 & x^2 \\ y^3 & z^3 & x^3 \\ 1 & 1 & 1 \end{vmatrix}$. Hence proved. Evaluation: in $\Delta = \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix}$ apply $R_1 \to R_1 - R_2$, $R_2 \to R_2 - R_3$ and take out $(x-y)(y-z)$; then $R_2 \to R_2 - R_1$ and take out $(z-x)$. Expanding along $C_3$ gives $\Delta = (x-y)(y-z)(z-x)(xy + yz + zx)$. Since $xy + yz + zx = 1$: $\Delta = (x-y)(y-z)(z-x)$.