Determinant of a specific $4 \times 4$ symmetric matrix In a recent research work, I have come across the following nice identity, where the entries $a,b,x$ belong to an arbitrary commutative unital ring:
$$\begin{vmatrix}
2 & a & b & ab-x \\
a & 2 & x & b \\
b & x & 2 & a \\
ab-x & b & a & 2 
\end{vmatrix}=(x^2-abx+a^2+b^2-4)^2.$$ 
Note that if the ring has characteristic $2$ then the formula is an obvious application of the Pfaffian.  
The only way I have been able to check this identity is through a tedious computation (of course, running any formal computing software will do). 
My question: Is there any elegant way to prove it?
 A: The following answer is inspired by Colin's.
As noted by Colin MacQuillan, the matrix under consideration turns out to be similar to
the block matrix 
$\begin{bmatrix}
A & C \\
C & B
\end{bmatrix}$
where $A=2I_2+x K$, $C=a I_2+ b K$, $B=2 I_2+(ab-x) K$, with
$K:=\begin{bmatrix}
0 & 1 \\
1 & 0 
\end{bmatrix}$. Since $I_2$ and $K$ commute, it is then standard that 
$$\begin{vmatrix}
A & C \\
C & B
\end{vmatrix}=\det(AB-C^2).$$
Here, we see that $AB=(4+x(ab-x))I_2+2ab K$ and $C^2=(a^2+b^2) I_2+2ab K$.
Hence, $AB-C^2=(4-a^2-b^2+x(ab-x))I_2$, which yields the claimed result.
Generalizing the diagonal entry of the initial matrix to a $c$ gives, with a similar method, that the determinant equals that of 
$(c^2-a^2-b^2+x(ab-x))I_2+(c-2)ab K$, yielding Andreas's formula. 
A: In short: determinant of every symmetric matrix is a square!
Consider arbitrary three matrices $a, b, c \in Sl_2.$ One can wonder, what relations do the traces of their products satisfy. The answer is given by the famous Jimbo-Fricke cubic:
\begin{equation}
 \begin{split}
&tr(ab)\  tr(bc) \ tr(ac)+ tr(ab)^2+tr(bc)^2+tr(ac)^2\\
&+tr(a)^2+tr(b)^2+tr(c)^2+tr(abc)^2\\
&-(tr(a)tr(b)+tr(c)tr(abc))tr(ab)\\
&-(tr(b)tr(c)+tr(a)tr(abc))tr(bc)\\
&-(tr(a)tr(c)+tr(b)tr(abc))tr(ac)\\
&+tr(a)tr(b)tr(c)tr(abc)-4=0.\\
 \end{split}
\end{equation}
Every determinant of a symmetric matrix can be written in the following form for some matrices $a, b, c$:
$$
G=\begin{vmatrix}
2 & -tr(a) & -tr(b) & -tr(bc)\\
-tr(a) & 2 & -tr(ab)& -tr(abc)\\
-tr(b) & -tr(ab) & 2 & -tr(c)\\
-tr(bc) & -tr(abc) & -tr(c)& 2\\
\end{vmatrix}.
$$
The relation above is equivalent to the following:
$$(2tr(ac)+tr(ab)tr(bc)-tr(a)tr(c)-tr(b)tr(abc))^2=G.$$
Usually a symmetric determinant is not a square, because $tr(ac)$ is not a polynomial in the entries of $G.$ The case of the matrix in the question corresponds to $c=a^{-1},$ because $tr(c)=tr(c^{-1}),$ $tr(aba^{-1})=tr(a)$ and $tr(ab)+tr(ba^{-1})=tr(a)+tr(b).$ The square root of $G$ is algebraic, because $tr(ac)=2.$
I have seen this presentation of the Jimbo-Fricke cubic only in one place: https://arxiv.org/pdf/1308.4092.pdf, formula (3.9), and I will be really grateful for any references.
A: Here's a method for calculating the determinant, explaining at least why it ends up as a product. I don't know if there's any significance to your determinant being a square.
Define
$$H=
\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & -1 \\
\end{pmatrix}.
$$
(The tensor product of a one-dimensional Hadamard matrix with a two-by-two identity matrix.)
Then $\det H=1$ and for any $a,b,c,d,e,f,g,h$,
$$
H
\begin{pmatrix}
a & b & c & d \\
b & a & d & c \\
e & f & g & h \\
f & e & h & g 
\end{pmatrix}
H\\
=\begin{pmatrix}
a+b & 0 & c+d & 0 \\
0 & a-b & 0 & c-d \\
e+f & 0 & g+h & 0 \\
0 & e-f & 0 & g-h 
\end{pmatrix}$$
which is (similar to)
$$\begin{pmatrix}
a+b & c+d  \\
e+f & g+h  \\
\end{pmatrix}
\oplus
\begin{pmatrix}
a-b & c-d  \\
e-f & g-h  \\
\end{pmatrix}.
$$ 
Plugging in a rotated version of your matrix gives
$$\begin{vmatrix}
2 & x & b & a \\
x & 2 & a & b \\
b & a & 2 & ab-x \\
a & b & ab-x & 2 
\end{vmatrix}
\\=
\begin{vmatrix}
2+x & a+b  \\
a+b & 2+ab-x  \\
\end{vmatrix}
\cdot
\begin{vmatrix}
2-x & b-a  \\
b-a & 2-ab+x  \\
\end{vmatrix}
\\
=(4-x^2+abx-a^2-b^2)(4-x^2+abx-a^2-b^2).
$$
