Trace of product of three Pauli matrices Consider the four $2\times 2$ matrices $\{\sigma_\mu\}$, with $\mu = 0,1,2,3$, which are defined as follows
$$
\sigma_0 =\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)
$$
$$
\sigma_1 =\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)
$$
$$
\sigma_2 =\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)
$$
$$
\sigma_3 =\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)
$$
i.e. the identity matrix and the three Pauli matrices. For the trace of the product of any two matrices $\sigma_\mu$ one has the identity $\text{tr}(\sigma_\mu \sigma_\nu)= 2 \delta_{\mu \nu}$. I was wondering if a similar identity can be derived for the product of three sigma matrices, 
$$
\text{tr}(\sigma_\mu \sigma_\nu \sigma_\lambda)= \;?
$$

First I note a few things about the defined matrices:
1) Commutation: $[\sigma_\mu,\sigma_\nu] = 2 i \epsilon_{0 \mu \nu \rho} \sigma_\rho$
2) Anti-commutation: $\{ \sigma_\mu ,\sigma_\nu \} = 2 \delta_{\{\mu \nu}  \sigma_{0\}} - 4 \delta_{\mu 0} \delta_{\nu 0} \sigma_0$
Where $a_{ \{b c} d_{i\}} = a_{bc} d_i + a_{ib} d_{c} + a_{ci}d_b$ 
This gives: $\sigma_\mu \sigma_\nu = i \epsilon_{0 \mu \nu \rho} \sigma_\rho + \delta_{\{\mu \nu} \sigma_{0\}}-\delta_{\mu 0} \delta_{\nu 0} \sigma_0$
Thus:
\begin{eqnarray}
\text{tr}(\sigma_\mu \sigma_\nu \sigma_\rho) &=& \text{tr}(i \epsilon_{0 \mu \nu \alpha} \sigma_\alpha\sigma_\rho + \delta_{\{\mu \nu} \sigma_{0\}}\sigma_\rho-2\delta_{\mu 0} \delta_{\nu 0} \sigma_\rho)\\
&=& 2 i \epsilon_{0\mu\nu\rho}+2\delta_{\{\mu \nu}\delta_{0\}\rho}-4\delta_{\mu0}\delta_{\nu 0} \delta_{\rho 0 }
\end{eqnarray}
where I have used the trace identity sketched out in the problem statement.
We can check the results of this expression by confirming that if one of the three, say $\mu=0$ then we get back the original identity.
We can also look at what happens if $\mu ,\nu,\rho \ne0$. In this case we would get $2i\epsilon_{0\mu \nu \rho}$ this matches the other answer. 
