= Solution
{c}
First derive the contracted <gamma matrices> identities from the <Clifford algebra> in four dimensions. Moving the outer <gamma matrix> through two factors gives
$$
\gamma_\rho\gamma_\alpha\gamma_\beta\gamma^\rho=4g_{\alpha\beta}\mathbf1.
$$
Moving it through three factors, and using this two-factor result, gives
$$
\begin{aligned}
\gamma_\rho\gamma_\alpha\gamma_\beta\gamma_\gamma\gamma^\rho
&=2\gamma_\beta\gamma_\gamma\gamma_\alpha-4g_{\beta\gamma}\gamma_\alpha\\
&=-2\gamma_\gamma\gamma_\beta\gamma_\alpha.
\end{aligned}
$$
Consequently $\gamma^\nu\not b\gamma_\mu\not c\gamma_\nu=-2\not c\gamma_\mu\not b$ and $\gamma_\mu\not b\not d\gamma^\mu=4(b\cdot d)\mathbf1$. Apply these in succession:
$$
\begin{aligned}
T&=\operatorname{tr}(\not a\gamma^\nu\not b\gamma_\mu\not c\gamma_\nu\not d\gamma^\mu)\\
&=-2\operatorname{tr}(\not a\not c\gamma_\mu\not b\not d\gamma^\mu)\\
&=-8(b\cdot d)\operatorname{tr}(\not a\not c)
=\boxed{-32(a\cdot c)(b\cdot d)}.
\end{aligned}
$$
The final step uses $\operatorname{tr}(\gamma^\alpha\gamma^\beta)=4g^{\alpha\beta}$. This is a genuinely four-dimensional instance of the <gamma matrix trace identities>; using it unchanged inside a $D$-dimensional loop calculation would require separate justification.
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