Factorizing four generators into two commutators Let $t_a$ be generators of any given Lie algebra such that $[t_a, t_b]=iC^{c}_{ab}t_c$.
Let $A_{a\mu}$ be gauge bosons associated with this Lie algebra. Here $\mu$ is the spacetime index and $a$ is the gauge index.
Then I want to show that 
$$\{tr(t_a t_b t_c t_d)\}A_{a\mu}A_{b\nu}A_{c\rho}A_{d\sigma}(-2\eta^{\mu \rho}\eta^{\nu \sigma}+\eta^{\mu \nu}\eta^{\rho \sigma}+\eta^{\mu \sigma}\eta^{\nu \rho})=-tr([t_a,t_b][t_c,t_d])A_{a\mu}A_{b\nu}A_{c}^{\mu}A_{d}^{\nu}$$
holds.
I tried to use the cyclicity of trace but I cannot cleanly factorize the product $t_a t_b t_c t_d$ into two commutators...
Could anyone please help me? Actually this is from p.106 of Weinberg QFT volume 2. It is the equation (17.5.30).
 A: There must be an elegant symmetry way to show this, but I opt for a quick one.
You first rename the contracted group indices on the left hand side so as to reduce the gauge fields to the canonical form of the group projector you see on the right hand side,
$$
A_{a\mu}A_{b\nu}A_{c}^{\mu}A_{d}^{\nu}\equiv S((ac),(bd)).
$$
This tensor is symmetric in (ac) and also in (bd), so contracted on an expression symmetrizes these pairs of it. 
You may work out the metric contractions on the left hand side, then, and rename group indices to obtain, 
$$
2\operatorname{Tr}(t_a t_c t_b t_d    -t_a t_b t_c t_d  ) ~ S((ac),(bd)),
$$
which you may rewrite exploiting the symmetries of the projector S  and the cyclicity of the trace,  as
$$
2\operatorname{Tr}(t_c t_a t_b t_d    -t_a t_b t_c t_d  ) ~ S((ac),(bd))\\
=2\operatorname{Tr}(t_a t_b t_d t_c   -t_a t_b t_c t_d  ) ~ S((ac),(bd))\\
=2\operatorname{Tr}( t_a t_b [t_d,t_c] ) ~ S((ac),(bd))  \\
=2\operatorname{Tr}( t_c t_d [t_b,t_a] ) ~ S((ac),(bd)) . 
$$
So, effectively, S projects out the pieces of the Trace tensor involved here that are symmetric in both (ab) and (cd), leaving only the [ab], and [cd] pieces standing.  Analyzing the products of two ts as the average of a commutator and anticommutator yields
$$
-\operatorname{Tr}( [t_c ,t_d] [t_a,t_b] ) ~ S((ac),(bd)) ,
$$
which is your right hand side.
