I need to calculate this expression:  

$$Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5}) $$ 
I know that I can express this as:
$$ Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})=-4i(g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\mu\sigma}\epsilon^{\nu\rho\alpha\beta}-g^{\mu\alpha}\epsilon^{\nu\rho\sigma\beta}+g^{\mu\beta}\epsilon^{\nu\rho\sigma\alpha}+g^{\nu\rho}\epsilon^{\mu\sigma\alpha\beta}-g^{\nu\sigma}\epsilon^{\mu\rho\alpha\beta}+g^{\nu\alpha}\epsilon^{\mu\rho\sigma\beta}-g^{\nu\beta}\epsilon^{\mu\rho\sigma\alpha}+g^{\rho\sigma}\epsilon^{\mu\nu\alpha\beta}-g^{\rho\alpha}\epsilon^{\mu\nu\sigma\beta}+g^{\rho\beta}\epsilon^{\mu\nu\sigma\alpha}+g^{\sigma\alpha}\epsilon^{\mu\nu\rho\beta}-g^{\sigma\beta}\epsilon^{\mu\nu\rho\alpha}+g^{\alpha\beta}\epsilon^{\mu\nu\rho\sigma}) $$
So, some of this terms are the same and some vanish. My question is how to show that.  

I know that:  

$$Tr(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\alpha}\gamma^{\beta}\gamma^{5})=-4i(g^{\mu\nu}\epsilon^{\rho\sigma\alpha\beta}-g^{\mu\rho}\epsilon^{\nu\sigma\alpha\beta}+g^{\rho\nu}\epsilon^{\mu\sigma\alpha\beta}-g^{\alpha\beta}\epsilon^{\sigma\mu\nu\rho}+g^{\sigma\beta}\epsilon^{\alpha\mu\nu\rho}-g^{\sigma\alpha}\epsilon^{\beta\mu\nu\rho}) $$
So only six terms survive, but how?


**EDIT:**  

The gamma matrices are defined [here](https://en.wikipedia.org/wiki/Gamma_matrices), the metric tensor is
$$g=\begin{bmatrix}
 1 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0 \\
 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & -1 \\
\end{bmatrix}$$
and $\epsilon^{ijkl}$ is Levi Civita symbol in four dimensions



Trace of six gamma matrices
To get from your 15 term expression to your 6 term expression, you can use what is sometimes called the Schouten identity, $g^{\mu [ \nu} \epsilon^{\rho \sigma \alpha \beta]} = 0$. The square brackets denote antisymmetrization, and an antisymmetrization of 5 indices in 4 dimensions must vanish. There are 5 terms when expanded: $g^{\mu \nu} \epsilon^{\rho \sigma \alpha \beta} + g^{\mu \rho} \epsilon^{\sigma \alpha \beta \nu} + g^{\mu \sigma} \epsilon^{\alpha \beta \nu \rho} + g^{\mu \alpha} \epsilon^{\beta \nu \rho \sigma} + g^{\mu \beta} \epsilon^{\nu \rho \sigma \alpha} = 0$. Apply the Schouten identity three times using $g^{\mu [ \nu} \epsilon^{\rho \sigma \alpha \beta]} = 0$, $g^{\nu [ \rho} \epsilon^{\mu \sigma \alpha \beta]} = 0$ and $g^{\rho [ \mu} \epsilon^{\nu \sigma \alpha \beta]} = 0$. The 9 terms you want to cancel will turn into 6 terms, but they are opposite sign pairs of $g^{\mu \nu} \epsilon^{\rho \sigma \alpha \beta}$, $g^{\nu \rho} \epsilon^{\mu \sigma \alpha \beta}$, $g^{\rho \mu} \epsilon^{\nu \sigma \alpha \beta}$, so they cancel.


