Easier way to identify a 3 variable determinant as symmetric polynomial How do I identify a 3 variable determinant whether it is a symmetric polynomial or not without actually expanding it ?
Examples:
$$
\begin{vmatrix}
1+a^2&ab&ac\\
ab&1+b^2&bc\\
ca&cb&1+c^2\\
\end{vmatrix}=1+a^2+b^2+c^2
$$
$$\begin{vmatrix}
x+a&y&z\\
x&y+a&z\\
x&y&z+a\\
\end{vmatrix}=a^2(x+y+z+a),\\\begin{vmatrix}
x-y-z&2x&2x\\
2y&y-z-x&2y\\
2z&2z&z-x-y\\
\end{vmatrix}=(x+y+z)^3
$$
$$
\begin{vmatrix}
3x&-x+y&-x+z\\
-y+x&3y&-y+z\\
-z+x&-z+y&3z
\end{vmatrix}=3(x+y+z)(xy+yz+zx),
\begin{vmatrix}
(y+z)^2 & x^2 & x^2 \\
y^2 & (z+x)^2 & y^2 \\
z^2 & z^2 & (x+y)^2 \\
\end{vmatrix}=2xyz(x+y+z)^3
$$
$$
\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3
,
\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}=4xyz\\
\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2
$$
$$
\begin{vmatrix}
1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\\
\end{vmatrix}=(1+a^2+b^2)^3
$$
My Attempt:
Lets take the example $\Delta=\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}$.
$$
\Delta_{x\leftrightarrow y}=\begin{vmatrix}
x+z&z&x\\
z&z+y&y\\
x&y&x+y
\end{vmatrix}=-\begin{vmatrix}
z&z+y&y\\
x+z&z&x\\
x&y&x+y
\end{vmatrix}=\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}=\Delta
$$
$$
\Delta_{x\leftrightarrow z}=\begin{vmatrix}
y+x&x&y\\
x&z+x&z\\
y&z&z+y
\end{vmatrix}
=-\begin{vmatrix}
y&z&z+y\\
x&z+x&z\\
y+x&x&y\\
\end{vmatrix}=
\begin{vmatrix}
z+y&z&y\\
z&z+x&x\\
y&x&y+x\\
\end{vmatrix}=\Delta
$$
$$
\Delta_{y\leftrightarrow z}=\begin{vmatrix}
y+z&y&z\\
y&y+x&x\\
z&x&x+z
\end{vmatrix}
=-\begin{vmatrix}
y+z&y&z\\
z&x&x+z\\
y&y+x&x\\
\end{vmatrix}=
\begin{vmatrix}
z+y&z&y\\
z&z+x&x\\
y&x&y+x\\
\end{vmatrix}=\Delta
$$
Thus $\Delta(x,y,z)$ is a symmetric polynomial.
Is there a better and easier way to identify such determinants ?
My Observation:
It may be a stupid observation while looking for a shortcut to the mentioned problem. For each of the cases the product of diagonal terms and product of the rest are separately symmetric polynomials. If not possible then we will do row or column operation and check the same.  Can I use it somehow to identify the determinant ?
Example:
$$
\begin{vmatrix}
a^2&bc&ac+c^2\\
a^2+ab&b^2&ac\\
ab&b^2+bc&c^2\\
\end{vmatrix}=4a^2b^2c^2
$$
is a symmetric polynomial of degree 6.
Here,
Product of the diagonal terms, $a^2b^2c^2$ is a symmetric polynomial.
Product of the rest, 
$$
(a^2+ab)(ac+c^2)(b^2+bc).ab.bc.ca=a(a+b).b(b+c).c(a+c).a^2b^2c^2\\
=a^3b^3c^3.(a+b)(b+c)(c+a)
$$
which is also a symmetric polynomial.
Note:
The reason why I am asking this is that If I can quickly identify by just looking at it using some shortcuts rather than actually checking for each cases, atleast for a $3x3$ matrix with 3 variables, it'd help ease to factorize the determinant using factor theorem. For example, check 
Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$
 A: I don't think there's a systematic answer. But a few observations.
(A) I think it is obvious that the examples you give satisfy the $x\to y\to z\to x$ symmetry, so you only need check that swapping $x$ and $y$ is a symmetry. Or if you don't want to do that, ask what would happen if swapping $x$ and $y$ were not a symmetry: in that case we'd  need to have a factor $(x-y)(y-z)(z-x)$, wouldn't we?[proof omitted]. Given the small degree of the determinant that's not going to be possible. 
(B) Most of these specific examples are determinants of matrices which can be written as $\alpha I + \beta J +\gamma J^2 + D$, where $J$ is the circulant matrix and $D$ is diagonal. In this form it is easier (at least for me) to check the symmetry. (For any $M$, $\det M =\det M \det J$.)
(C) The first case you give is actually easier to tackle by regarding $a$ as the variable: it's trivial to compute the coefficients of $a^3, a^2, a, 1$. This is often a good way to tackle a determinant of the form $A+t I$ where $A$ is free of $t$. 
(D) The example with the $yz-x^2$ entries is easier (as I remarked on your previous question) if we spot that it is a matrix of cofactors and so a perfect square.
