If are distinct non zero real numbers, prove that . Hence or otherwise evaluate in its simplest…

Given $\Delta = \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix}$. Multiplying $R_1$ by $x$, $R_2$ by $y$, and $R_3$ by $z$: $\Delta = \frac{1}{xyz} \begin{vmatrix} x^2 & x^3 & xyz \\ y^2 & y^3 & xyz \\ z^2 & z^3 & xyz \end{vmatrix}$. Factoring out $xyz$ from $C_3$: $\Delta = \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix}$. Using the determinant property $|A| = |A^T|$ and swapping rows: Interchanging $R_1$ and $R_2$, then $R_2$ and $R_3$ gives: $\Delta = \begin{vmatrix} y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \\ x^2 & x^3 & 1 \end{vmatrix}$. Hence proved. To evaluate $\Delta$, apply row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ on $\begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix}$: $\Delta = \begin{vmatrix} x^2 - y^2 & x^3 - y^3 & 0 \\ y^2 - z^2 & y^3 - z^3 & 0 \\ z^2 & z^3 & 1 \end{vmatrix} = (x - y)(y - z) \begin{vmatrix} x+y & x^2+xy+y^2 & 0 \\ y+z & y^2+yz+z^2 & 0 \\ z^2 & z^3 & 1 \end{vmatrix}$. Applying $R_2 \to R_2 - R_1$: $= (x - y)(y - z) \begin{vmatrix} x+y & x^2+xy+y^2 & 0 \\ z-x & (z-x)(x+y+z) & 0 \\ z^2 & z^3 & 1 \end{vmatrix} = (x - y)(y - z)(z - x) \begin{vmatrix} x+y & x^2+xy+y^2 \\ -1 & -(x+y+z) \end{vmatrix}$ $= (x - y)(y - z)(z - x) [-(x+y)(x+y+z) + (x^2+xy+y^2)]$ $= (x - y)(y - z)(z - x) [-(x^2 + 2xy + y^2 + xz + yz) + x^2 + xy + y^2]$ $= (x - y)(y - z)(z - x) [-(xy + yz + zx)]$. Since $xy + yz + zx = 1$, we get: $\Delta = -(x - y)(y - z)(z - x) = (x - y)(y - z)(x - z)$.