Show that $\Delta=\left|\begin{array}{ccc} (y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & y z & (x+y)^{2} \end{array}\right|=2 x y z(x+y+z)^{3}$

(N/A) Applying $R_{1} \rightarrow x R_{1}, R_{2} \rightarrow y R_{2}, R_{3} \rightarrow z R_{3}$ to $\Delta$ and dividing by $x y z,$ we get
$\Delta=\frac{1}{x y z}\left|\begin{array}{ccc} x(y+z)^{2} & x^{2} y & x^{2} z \\ x y^{2} & y(x+z)^{2} & y^{2} z \\ x z^{2} & y z^{2} & z(x+y)^{2} \end{array}\right|$
Taking common factors $x, y, z$ from $C_{1}, C_{2}$ and $C_{3}$ respectively,we get
$\Delta=\frac{x y z}{x y z}\left|\begin{array}{ccc} (y+z)^{2} & x^{2} & x^{2} \\ y^{2} & (x+z)^{2} & y^{2} \\ z^{2} & z^{2} & (x+y)^{2} \end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-C_{1},$ we have
$\Delta=\left|\begin{array}{ccc} (y+z)^{2} & x^{2}-(y+z)^{2} & x^{2}-(y+z)^{2} \\ y^{2} & (x+z)^{2}-y^{2} & 0 \\ z^{2} & 0 & (x+y)^{2}-z^{2} \end{array}\right|$
Taking common factor $(x+y+z)$ from $C_{2}$ and $C_{3},$ we have
$\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} (y+z)^{2} & x-(y+z) & x-(y+z) \\ y^{2} & (x+z)-y & 0 \\ z^{2} & 0 & (x+y)-z \end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-(R_{2}+R_{3}),$ we have
$\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} 2 y z & -2 z & -2 y \\ y^{2} & x-y+z & 0 \\ z^{2} & 0 & x+y-z \end{array}\right|$
Applying $C_{2} \rightarrow (C_{2}+\frac{1}{y} C_{1})$ and $C_{3} \rightarrow C_{3}+\frac{1}{z} C_{1},$ we get
$\Delta=(x+y+z)^{2}\left|\begin{array}{ccc} 2 y z & 0 & 0 \\ y^{2} & x+z & \frac{y^{2}}{z} \\ z^{2} & \frac{z^{2}}{y} & x+y \end{array}\right|$
Finally expanding along $R_{1},$ we have
$\Delta = (x+y+z)^{2}(2 y z)[(x+z)(x+y)-y z]$
$= (x+y+z)^{2}(2 y z)(x^{2}+x y+x z)$
$= (x+y+z)^{3}(2 x y z)$