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?

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
$$\begin{vmatrix}
2 & x & b & a \\
x & 2 & a & b \\
b & a & 2 & ab-x \\
a & b & ab-x & 2 
\end{vmatrix}
\\=
\begin{vmatrix}
2+x & a+b  \\
a+b & 2+ab-x  \\
\end{vmatrix}
\cdot
\begin{vmatrix}
2-x & b-a  \\
b-a & 2-ab+x  \\
\end{vmatrix}
\\
=(4-x^2+abx-a^2-b^2)(4-x^2+abx-a^2-b^2).
$$
