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.