Using properties of determinants, prove that \begin{vmatrix} (x+y)^2 & zx & zy \ zx & (z+y)^2 & xy \ zy & xy & (z+x)^2 \end{vmatrix} = 2xyz(x+y+z)^3

\begin{vmatrix} (x+y)^2 & zx & zy \ zx & (z+y)^2 & xy \ zy & xy & (z+x)^2 \end{vmatrix} \R1 \rightarrow zR1, R2 \rightarrow xR2, R3 \rightarrow yR3 \we \ get = \frac{1}{xyz} \begin{vmatrix} z(x+y)^2 & z^2x & z^2y \ x^2z & x(z+y)^2 & x^2y \ y^2z & y^2x & y(z+x)^2 \end{vmatrix} \Taking \ z, \ x, \ y \ common \ from \ C1, C2, C3 \ respectively, = \frac{xyz}{xyz} \begin{vmatrix} (x+y)^2 & z^2 & z^2 \ x^2 & (z+y)^2 & x^2 \ y^2 & x^2 & (z+x)^2 \end{vmatrix} \C1 \rightarrow C1 - C3, C2 \rightarrow C2 - C3 = \begin{vmatrix} (x+y)^2 - z^2 & 0 & z^2 \ x^2 - x^2 & (z+y)^2 - x^2 & x^2 \ y^2 - (z+x)^2 & x^2 - (z+x)^2 & (z+x)^2 \end{vmatrix} \ \rightarrow \begin{vmatrix} (x+y+z)(x+y-z) & 0 & z^2 \ 0 & (y+z-x)(y+z+x) & x^2 \ (y-z-x)(y+z+x) & (x-z-y)(x+z+y) & (z+x)^2 \end{vmatrix} \ \rightarrow (x+y+z)^2 \begin{vmatrix} x+y-z & 0 & z^2 \ 0 & y+z-x & x^2 \ y-z-x & x-z-y & (z+x)^2 \end{vmatrix} \R3 \rightarrow R3 - R2 - R1 \rightarrow (x+y+z)^2 \begin{vmatrix} x+y-z & 0 & z^2 \ 0 & y+z-x & x^2 \ -2x & -2z & 2zx \end{vmatrix} \C1 \rightarrow C1 + \frac{1}{z}C3, C2 \rightarrow C2 + \frac{1}{x}C3 \rightarrow (x+y+z)^2 \begin{vmatrix} x+y & z^2/x+2z \ z^2/z + 2z & y+z \ 0 & 0 \ 0 & 0 \end{vmatrix} \Expanding \ along \ R3 \rightarrow 2xz(x+y+z)^2((x+y)(z+y) - x^2z.z^2x) \rightarrow 2xz(x+y+z)^2(xz+xy+yz+y^2 - xz) \rightarrow 2xyz(x+y+z)^3 = RHS \Hence \ proved.