Trace of product of commutators in $\operatorname{SL}(2,\mathbb{C})$ Let $A,B\in\operatorname{SL}(2,\mathbb{C})$ and define $x=\operatorname{tr}(A)$, $y=\operatorname{tr}(B)$, $z=\operatorname{tr}(AB)$. It is know that these satisfy the equation $$\operatorname{tr}([A,B])=x^2+y^2+z^2-xyz-2,$$
where the commutator $[A,B]=ABA^{-1}B^{-1}$ is defined in the group setting. 
For $A,B,C,D\in\operatorname{SL}(2,\mathbb{C})$ can one obtain a similar formula for $\operatorname{tr}([A,B][C,D])$ in terms of $x=\operatorname{tr}(A)$, $y=\operatorname{tr}(B)$, $z=\operatorname{tr}(AB)$, $r=\operatorname{tr}(C)$, $s=\operatorname{tr}(D)$, $t=\operatorname{tr}(CD)$?
Edit:
I think such an equation will also need to include the variables $u=\operatorname{tr}(AC)$, $v=\operatorname{tr}(BD)$, $w=\operatorname{tr}(AD)$, $\omega=\operatorname{tr}(BC)$ (so that its level sets are $8$ dimensional surfaces).
 A: More of an observation:
A bit more general identity is true for $2\times 2$ matrices $A$, $B$
$$\operatorname{tr} (A B \operatorname{adj}(A)\operatorname{adj}(B))+ \operatorname{tr} (A) \operatorname{tr} (B) \operatorname{tr} (A B) - \operatorname{tr} (A B)^2 = $$
$$=\operatorname{tr} (A)^2 \det (B) + \operatorname{tr} (B)^2 \det (A) - 2\det (A) \det(B)
$$
where $\operatorname{adj}(A)$ is the adjugate
of the matrix $A$. If $A$ has determinant $1$, then $\operatorname{adj}(A)= A^{-1}$.
For the check with WA see 1 and 2. It is an equality of two biquadratic forms on a space of dimension $8$.
I suggest looking at a similar equality for $A$, $B$, $C$, $D$ $\  2\times 2$ matrices where we substitute the adjugate for the inverse, and also involve determinants. Perhaps using Groebner bases would be helpful.
Could be connected with polynomial invariants of the group $SL(2, \mathbb{C})$ acting on uples of $2\times 2$ matrices by conjugation.
$\bf{Added:}$ I looked up the proof of the indentity of Fricke in the book quoted by the Wikipadia article in the comments above. It is based on the following identities:
$$A^2 - \operatorname{tr}(A)\cdot A + \det (A) I_2 = 0 $$
$$A + \operatorname{adj}(A)= \operatorname{tr}(A) \cdot I_2 $$
or, if one prefers, the identities for matrices of unit determinant. Worth redoing the proof, and then seeing what can be said in other cases.
