Question

$\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|=(x-y)(y-z)(z-x)(xy+yz+zx)$

Answer 1

$LHS=\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|=\frac{1}{xyz}\left|\begin{array}{ccc}{x}^2& x^3& xyz\\ y^2& y^3& xyz\\ z^2& z^3& xyz\end{array}\right|$
(using $(R_1\to x{R}_1$ and $R_2\to y{R}_2$ and $R_3\to z{R}_3$

$=\frac{xyz}{xyz}\left|\begin{array}{ccc}{x}^2& x^3& 1\\ y^2& y^3& 1\\ z^2& z^3& 1\end{array}\right|$ (take out xyz common from $C_3$)

$=\left|\begin{array}{ccc}{x}^2& x^3& 1\\ y^2-x^2& y^3-x^3& 0\\ z^2-x^2& z^3-x^3& 0\end{array}\right|$ using $R_2\to R_2-R_{1}and{R}_3\to R_3-R_1$)
Expanding corresponding to $C_3$. we get

$=1\left|\begin{array}{cc}{y}^2-x^2& y^3-x^3\\ z^2-x^2& z^3-x^3\end{array}\right|=\left[\right(y^2-x^2\left)\right(z^3-x^3)-(z^2-x^2\left)\right(y^3-x^3\left)\right]$

$=(y+x)(y-x)(z-x)(z^2+x^2+xz)-(z+x)(z-x)(y-x)(y^2+x^2+xy)=(y-x)(z-x)[y{z}^2+y{x}^2+xyz+x{z}^2+x^3+x^{2}z-z{y}^2-z{x}^2-xyz-x{y}^2-x^3-x^{2}y]$

$=(y-x)(z-x)[y{z}^2-z{y}^2+x{z}^2-x{y}^2]=(y-x)(z-x)\left[yz\right(z-y)+x(z^2-y^2\left)\right]=(y-x)(z-x)\left[yz\right(z-y)+x(z-y\left)\right(z=y\left)\right]=(y-x)(z-x)\left[\right(z-y\left)\right(xy+yz+zx\left)\right]=(x-y)(y-x)(z-x)(xy+yz+zx)=RHS$.