By using properties of determinants, show that:

Let $\Delta=\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$.

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:

$\begin{aligned} \Delta &=\left|\begin{array}{lcc}x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z\end{array}\right| \\ &=\left|\begin{array}{ccc}x & x^{2} & y z \\ -(x-y) & -(x-y)(x+y) & z(x-y) \\ (z-x) & (z-x)(z+x) & -y(z-x)\end{array}\right| \\ &=(x-y)(z-x)\left|\begin{array}{lll}x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & z+x & -y\end{array}\right| \end{aligned}$

Applying $R_{3} \rightarrow R_{3}+R_{2}$, we have:

$\begin{aligned} \Delta &=(x-y)(z-x)\left|\begin{array}{lll}x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & z-y & z-y\end{array}\right| \\ &=(x-y)(z-x)(z-y)\left|\begin{array}{llll}x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & 1 & 1\end{array}\right| \end{aligned}$

Expanding along R3, we have:

$\begin{aligned} \Delta &=[(x-y)(z-x)(z-y)]\left[(-1)\left|\begin{array}{ll}x & y z \\ -1 & z\end{array}\right|+1\left|\begin{array}{lc}x & x^{2} \\ -1 & -x-y\end{array}\right|\right.\\ &=(x-y)(z-x)(z-y)\left[(-x z-y z)+\left(-x^{2}-x y+x^{2}\right)\right] \\ &=-(x-y)(z-x)(z-y)(x y+y z+z x) \\ &=(x-y)(y-z)(z-x)(x y+y z+z x) \end{aligned}$

Hence, the given result is proved.