How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3}{=}\begin{vmatrix}
xy+y^2+yz+zx+yz+z^2&xy&zx\\
x^2+xy+xz+xz+yz+z^2&(x+z)^2&yz\\
x^2+xy+xz+xy+y^2+yz&yz&(x+y)^2\\
\end{vmatrix}\\
=\begin{vmatrix}
y(x+y+z)+z(x+y+z)&xy&zx\\
x(x+y+z)+z(x+y+z)&(x+z)^2&yz\\
x(x+y+z)+y(x+y+z)&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)\begin{vmatrix}
y+z&xy&zx\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
\stackrel{R_1\rightarrow R_1+R_2+R_3}{=} \\(x+y+z)\begin{vmatrix}
2(x+y+z)&x^2+xy+xz+xz+yz+z^2&x^2+xy+zx+xy+y^2+yz\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)^2\begin{vmatrix}
2&x+z&x+y\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2
\end{vmatrix}
\stackrel{R_2\rightarrow R_2-(x+z)R_1|R_2\rightarrow R_3-(x+y)R_1}{=}
2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
$$
How do I extract $x+y+z$ or $xyz$ from the determinant to find the solution ?
Note: In a similar problem How to solve this determinant there seems to be solutions talking about factor theorem and polynomials. I am basically looking for extracting the terms $(x+y+z)$ or $xyz$ from the given determinant to solve it only using the basic properties of determinants.

It is easy to calculate
\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
If you still want to use extract $x+y+z$, you can use
\begin{eqnarray}
\begin{vmatrix}
\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{(x+y)^2}{2}-\frac{(yz-x^2-xy-xz)^2}{2(x+z)^2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{(x+y)^2(x+z)^2-(yz-x^2-xy-xz)^2}{2(x+z)^2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{xyz(x+y+z)}{2(x+z)^2}
\end{vmatrix}\\
&=&xyz(x+y+z)\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&1
\end{vmatrix}.
\end{eqnarray}
