Show that $ \text{Tr} ( XYXZ) = \text{Tr}(XY) \text{Tr}(XZ) - \text{Tr}( YZ^{-1} ) $ For $X,Y,Z \in \text{SL}_2(\mathbb{C})$ we have the following identities for $2 \times 2$ matrices: 
$$ \text{Tr} \big( XYXZ\big) = \text{Tr}\big(XY\big) \text{Tr}\big(XZ\big) - \text{Tr}\big( YZ^{-1} \big) \tag{$\ast$}$$
These are stated without proof, but they do give the following hint: basically the Cayley-Hamilton identity
$$ X^2 = \text{Tr}(X)\, X -I$$
and many other interesting formulas on that page.  
Underlying all of that... are all these $2 \times 2$ matrix identities fundamentally different from things like $\text{Tr}(I) = 2$ and $\text{Tr}(XY) = \text{Tr}(YX) $.  This equation is taken from a book on quaternions.

Here are some thought's using Dirac's bra-ket notation (rather than indices):
\begin{eqnarray*} \text{Tr}(XY) &=& \langle 1 | XY | 1 \rangle + \langle 2 | XY | 2 \rangle \\ \\
&=& 
\Big( \langle 1 | X | 1 \rangle \langle 1 | Y | 1 \rangle +
\langle 1 | X | 2 \rangle \langle 2 | Y | 1 \rangle \Big) +
\Big( \langle 2 | X | 1 \rangle \langle 1 | Y | 2 \rangle +
\langle 2 | X | 2 \rangle \langle 2 | Y | 2 \rangle \Big) \\ \\
&=& 
\Big( \langle 1 | X | 1 \rangle \langle 1 | Y | 1 \rangle +
\langle 2 | X | 1 \rangle \langle 1 | Y | 2 \rangle \Big) +
\Big( \langle 1 | X | 2 \rangle \langle 2 | Y | 1 \rangle +
\langle 2 | X | 2 \rangle \langle 2 | Y | 2 \rangle \Big) \\ \\
&=& \text{Tr}(YX) \end{eqnarray*}
Even an index proof could be somewhat informative here.
 A: First note that
$$ \text{Tr} \big( XYXZ\big) =\text{Tr} \big( XZXY\big) =\text{Tr} \big( ZXZXYZ^{-1}\big) =\text{Tr} \big( (ZX)^2YZ^{-1}\big). $$
By using 
$$ X^2 = \text{Tr}(X)\, X -I$$
one has
\begin{eqnarray}
\text{Tr} \big( XYXZ\big)&=&\text{Tr} \big( (ZX)^2YZ^{-1}\big)\\
&=&\text{Tr} \big( (\text{Tr}(ZX)ZX-I)YZ^{-1}\big)\\
&=&\text{Tr} \big(\text{Tr}(ZX)ZXYZ^{-1}-YZ^{-1}\big)\\
&=&\text{Tr}(ZX)\text{Tr} \big(ZXYZ^{-1}\big)-\text{Tr}\big(YZ^{-1}\big)\\
&=&\text{Tr}(ZX)\text{Tr} \big(XY\big)-\text{Tr}\big(YZ^{-1}\big)\\
&=&\text{Tr}(XY)\text{Tr} \big(XZ\big)-\text{Tr}\big(YZ^{-1}\big).
\end{eqnarray}
A: The claim works in any field.  Hence, I shall prove that
$$\text{Tr}(XYXZ)=\text{Tr}(XY)\,\text{Tr}(XZ)-\text{Tr}(YZ^{-1})$$
for all $X,Y,Z\in\text{SL}_2(\mathbb{K})$ for an arbitrary field $\mathbb{K}$.  For convenience, $0$ also denotes the $2$-by-$2$ zero matrix and $I$ is the $2$-by-$2$ identity matrix.
We have by the Cayley-Hamilton Theorem that $$T^2-\text{Tr}(T)\,T+\det(T)\,\text{I}=0$$ for all matrices $T\in\text{Mat}_{2\times 2}(\mathbb{K})$.  With $T:=XZ$ (noting that $\det(X)=\det(Z)=1$), we obtain
$$XZ=\text{Tr}(XZ)\,I-(XZ)^{-1}=\text{Tr}(XZ)\,I-Z^{-1}X^{-1}\,.$$
That is,
$$XYXZ=\text{Tr}(XZ)\,XY-XYZ^{-1}X^{-1}\,.$$
Taking the trace, we get
$$\text{Tr}(XYXZ)=\text{Tr}(XZ)\,\text{Tr}(XY)-\text{Tr}(XYZ^{-1}X^{-1})\,.$$
That is,
$$\text{Tr}(XYXZ)=\text{Tr}(XZ)\,\text{Tr}(XY)-\text{Tr}(YZ^{-1}X^{-1}X)=\text{Tr}(XY)\,\text{Tr}(XZ)-\text{Tr}(YZ^{-1})\,.$$
