By using properties of determinants,show that:
$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$

(N/A) 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 get:
$\Delta=\left|\begin{array}{ccc}x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z\end{array}\right|$
Factoring out $(y-x)$ from $R_{2}$ and $(z-x)$ from $R_{3}$:
$\Delta=(y-x)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -(x+y) & z \\ 1 & z+x & -y\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}+R_{2}$:
$\Delta=(y-x)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -(x+y) & z \\ 0 & z-y & z-y\end{array}\right|$
Taking $(z-y)$ common from $R_{3}$:
$\Delta=(y-x)(z-x)(z-y)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -(x+y) & z \\ 0 & 1 & 1\end{array}\right|$
Expanding along $R_{3}$:
$\Delta=(y-x)(z-x)(z-y) [0 - 1(x z + y z) + 1(-x^{2}-x y + x^{2})]$
$\Delta=(y-x)(z-x)(z-y) [-(x z + y z) - x y]$
$\Delta=-(y-x)(z-x)(z-y)(x y+y z+z x)$
Since $-(y-x) = (x-y)$ and $-(z-y) = (y-z)$,we get:
$\Delta=(x-y)(y-z)(z-x)(x y+y z+z x)$.
Hence,the result is proved.