Show that

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

Applying $\mathrm{R}_{1} \rightarrow x \mathrm{R}_{1}, \mathrm{R}_{2} \rightarrow y \mathrm{R}_{2}, \mathrm{R}_{3} \rightarrow z \mathrm{R}_{3}$ 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 $\mathrm{C}_{1} \mathrm{C}_{2}$ and $\mathrm{C}_{3},$ 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 $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1},$ 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 $\mathrm{C}_{2}$ and $\mathrm{C}_{3},$ 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 $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\left(\mathrm{R}_{2}+\mathrm{R}_{3}\right),$ 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 $\mathrm{C}_{2} \rightarrow\left(\mathrm{C}_{2}+\frac{1}{y} \mathrm{C}_{1}\right)$ and $\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}+\frac{1}{z} \mathrm{C}_{1},$ 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 $\mathrm{R}_{1},$ we have

$\Delta $$=(x+y+z)^{2}(2 y z)[(x+z)(x+y)-y z]$

$=(x+y+z)^{2}(2 y z)\left(x^{2}+x y+x z\right) $
$=(x+y+z)^{3}(2 x y z)$