Showing dual field strength tensor term in Lagrangian is a total derivative I am trying to prove the identity:
$$\text{Tr}\left\{\star F_{\mu\nu}F^{\mu\nu}\right\}=\partial^{\mu}K_{\mu} \tag{1}$$
where $K_{\mu}$ is given by:
$$K_\mu=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{A^{\nu}F^{\rho\sigma}-\frac{2}{3}igA^{\nu}A^{\rho}A^{\sigma}\right\}\tag{2}$$
and $F$ and $\star F$ are the (non-abelian) field strength tensors defined by:
$$\begin{align}
F^{\mu\nu}&\equiv\partial^{\mu}A^{\nu}-\partial^{\mu}A^{\nu}+ig[A^{mu},A^{\nu}]\\
&=D^{\mu}A^\nu-D^{\nu}A^{\mu},
\end{align}\tag{3}$$
$$\star F^{\mu\nu}\equiv \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}.\tag{4}$$
I have already seen this directly related question but after following the advice given there (namely expand the RHS of eq. 1 and use both the Bianchi identity for $F$ and dummy index relabeling) I still cannot prove the identity. Below is my work.

$$\begin{align*}
&\partial^{\mu}\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{A^{\nu}F^{\rho\sigma}-\frac{2}{3}igA^{\nu}A^{\rho}A^{\sigma}\right\}\\
&=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})F^{\rho\sigma}+\underset{\text{Bianchi}}{\underbrace{A^{\nu}\require{cancel}\cancel{(\partial^{\mu}F^{\rho\sigma})}}}-\underset{\text{index relabeling + cyclic perm.}}{\underbrace{2ig(\partial^{\mu}A^{\nu})A^{\rho}A^{\sigma}}}\right\} \tag{5}\\
&\\
&=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})F^{\rho\sigma}-ig(\partial^{\mu}A^{\nu})[A^{\rho},A^{\sigma}]\right\}\tag{6}\\
&=\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{(\partial^{\mu}A^{\nu})(F^{\rho\sigma}-ig[A^{\rho},A^{\sigma}])\right\}\tag{7}\\
&=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}\text{Tr}\left\{F^{\mu\nu}F^{\rho\sigma}\color{red}{-2igF^{\mu\nu}[A^{\rho},A^{\sigma}]-g^2[A^{\mu},A^{\nu}][A^{\rho},A^{\sigma}]}\right\}\tag{8}\\
\end{align*}$$
In order for the identity to be true the terms in red must vanish, which I cannot see to be true in general. Where have I gone wrong?
 A: As alluded to in the comments, using forms is much easier here. We note the formulae,
$$(\ast F)_{\mu\nu} F^{\mu\nu}  = \ast \text{tr}[F \wedge F],\qquad F = d A + A \wedge A,\qquad dF + A \wedge F - F \wedge A = 0 .$$
The quantity $\ast K = \text{tr} [ A \wedge F - \frac{1}{3} A \wedge A \wedge A ]$ and $\nabla_\mu K^\mu = \ast d \ast K$. We are now ready to prove our result, which now reads
$$
\text{tr}[ F \wedge F ] = d \ast K .
$$
Start with the RHS
\begin{align}
d \ast K &= \text{tr}[d A \wedge F - A \wedge d F - d A \wedge A \wedge A]  \\
&= \text{tr}[(F - A \wedge A ) \wedge F - A \wedge ( - A \wedge F + F \wedge A )   - (F - A \wedge A ) \wedge A \wedge A] \\
&= \text{tr}[F \wedge F - A \wedge A \wedge F + A \wedge A \wedge F- A \wedge F \wedge A - F \wedge A \wedge A + A \wedge A \wedge A \wedge  A ]\\
&= \text{tr}[F \wedge F + A \wedge A \wedge A \wedge  A ] . \\
\end{align}
where we used the fact $\text{tr}[A \wedge F \wedge A + F \wedge A \wedge A] = 0 $ (verify!)
The final thing we have to show is that $\text{tr}[A \wedge A \wedge A \wedge  A ] =0$. To see this, let us put in the explicit Lie algebra indices
\begin{align}
\text{tr}[A \wedge A \wedge A \wedge  A ] &= \text{tr}[T_a T_b T_cT_d] A^a \wedge A^b \wedge A^c \wedge A^d \\
&= \text{tr}[ [ T_a ,  T_b ] [ T_c , T_d ] ] A^a \wedge A^b \wedge A^c \wedge A^d \\
&=f_{abe} f_{ecd} A^a \wedge A^b \wedge A^c \wedge A^d \\
&= - ( f_{cae} f_{ebd} + f_{bce} f_{ead} ) A^a \wedge A^b \wedge A^c \wedge A^d \\
&=- 2 f_{abe} f_{ecd}  A^a \wedge A^b \wedge A^c \wedge A^d.
\end{align}
Here, we used the Jacobi identity and then did some index manipulation in the last step (do this yourself!). Thus, $\text{tr}[A \wedge A \wedge A \wedge  A ]  = 0$. 
Hence
$$
\text{tr} [ F \wedge F ] = d \ast  K.
$$
