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.
 A: The determinant must be a polynomial in $x,y,z$ of degree $6$.
If you set $x=0$,
$$
\begin{vmatrix}
(y+z)^2&0&0\\
0&z^2&yz\\
0&yz&y^2
\end{vmatrix}=0
$$
so that $x$ is a factor. And by symmetry, $xyz$ as well.
Then with $x=-y-z$,
$$
\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&y^2&yz\\
xz&yz&z^2
\end{vmatrix}=0
$$ (the bottom six minors are zero), and $x+y+z$ is a factor.
To reach the sixth degree, we are missing two other linear factors. But by symmetry, they cannot be other than $x+y+z$ both (two factors alone cannot sustain a permutation of the variables).
Remains to find the global factor, for example from
$$
\begin{vmatrix}
4&1&1\\
1&4&1\\
1&1&4
\end{vmatrix}=\lambda\cdot1\cdot1\cdot1\cdot(1+1+1)^3.$$
A: 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}
A: Let $s=x+y+z$, then
$$
\begin{align}
&\det\begin{bmatrix}
(y+z)^2&xy&zx\\
xy&(z+x)^2&yz\\
zx&yz&(x+y)^2
\end{bmatrix}\\[9pt]
&=\det\begin{bmatrix}
s(y+z)&xy&zx\\
s(z+x)&(z+x)^2&yz\\
s(x+y)&yz&(x+y)^2
\end{bmatrix}\tag1\\[9pt]
&=\det\begin{bmatrix}
2s^2&s(z+x)&s(x+y)\\
s(z+x)&(z+x)^2&yz\\
s(x+y)&yz&(x+y)^2
\end{bmatrix}\tag2\\[9pt]
&=s^2\det\begin{bmatrix}
2&z+x&x+y\\
z+x&(z+x)^2&yz\\
x+y&yz&(x+y)^2
\end{bmatrix}\tag3\\[9pt]
&=s^2\det\begin{bmatrix}
2&-(z+x)&-(x+y)\\
z+x&0&yz-(x+y)(z+x)\\
x+y&yz-(x+y)(z+x)&0
\end{bmatrix}\tag4\\[9pt]
&=s^2\det\begin{bmatrix}
2&-(z+x)&-(x+y)\\
z+x&0&-sx\\
x+y&-sx&0
\end{bmatrix}\tag5\\[9pt]
&=s^3\det\begin{bmatrix}
2s&z+x&x+y\\
z+x&0&x\\
x+y&x&0
\end{bmatrix}\tag6\\[9pt]
&=s^3\det\begin{bmatrix}
y+z&z+x&x+y\\
z&0&x\\
y&x&0
\end{bmatrix}\tag7\\[9pt]
&=s^3\det\begin{bmatrix}
0&z&y\\
z&0&x\\
y&x&0
\end{bmatrix}\tag8\\[18pt]
&=2xyz(x+y+z)^3\tag9
\end{align}
$$
Explanation:
$(1)$: add columns $2$ and $3$ to column $1$
$(2)$: add rows $2$ and $3$ to row $1$
$(3)$: factor $s$ out of row $1$ and then out of column $1$
$(4)$: subtract $z+x$ times column $1$ from column $2$
$\phantom{(4)\text{:}}$ subtract $x+y$ times column $1$ from column $3$
$(5)$: $yz-(x+y)(z+x)=-sx$
$(6)$: factor $-s$ out of columns $2$ and $3$
$\phantom{(6)\text{:}}$ distribute one factor of $s$ over row $1$
$(7)$: subtract columns $2$ and $3$ from column $1$
$(8)$: subtract rows $2$ and $3$ from row $1$
$(9)$: the determinant is now simple to compute
A: $$
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=xyz\begin{vmatrix}
\frac{(y+z)^2}{x}&x&x\\
y&\frac{(x+z)^2}{y}&y\\
z&z&\frac{(x+y)^2}{z}
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{(y+z)^2}{x}-x&x&0\\
y-\frac{(x+z)^2}{y}&\frac{(x+z)^2}{y}&y-\frac{(x+z)^2}{y}\\
0&z&\frac{(x+y)^2}{z}-z
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{(y+z)^2-x^2}{x}&x&0\\
\frac{y^2-(x+z)^2}{y}&\frac{(x+z)^2}{y}&\frac{y^2-(x+z)^2}{y}\\
0&z&\frac{(x+y)^2}{z}
\end{vmatrix}=
xyz\begin{vmatrix}
\frac{[(y+z)-x][(y+z)+x]}{x}&x&0\\
\frac{[y-(x+z)][y+(x+z)]}{y}&\frac{(x+z)^2}{y}&\frac{[y-(x+z)][y+(x+z)]}{y}\\
0&z&\frac{[(x+y)-z][(x+y)+z]}{z}
\end{vmatrix}=
xyz(x+y+z)^2\begin{vmatrix}
\frac{[(y+z)-x]}{x}&x&0\\
\frac{[y-(x+z)]}{y}&\frac{(x+z)^2}{y}&\frac{[y-(x+z)]}{y}\\
0&z&\frac{[(x+y)-z]}{z}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
{[y-(x+z)]}&{(x+z)^2}&{[y-(x+z)]}\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}
=(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
{[y-(x+z)]}&{(x+z)^2}&{[y-(x+z)]}\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{[(y+z)-x]}&x^2&0\\
-2z&2xz&-2x\\
0&z^2&{[(x+y)-z]}
\end{vmatrix}=
(x+y+z)^2\begin{vmatrix}
{(y+z)}&x^2&\frac{x^2}{z}\\
0&2xz&0\\
\frac{z^2}{x}&z^2&{(x+y)}
\end{vmatrix}\\
=(x+y+z)^2.2xz.\Big[(x+y)(y+z)-\frac{x^2.z^2}{xz}\Big]\\
=(x+y+z)^2(2xz)(xy+xz+y^2+xz-xz)\\
=(x+y+z)^2(2xz)(xy+y^2+xz)\\=(x+y+z)^2.2xz.y(x+y+z)=2xyz(x+y+z)^2
$$
