= Solution
{c}
Use the <Feynman rules> and <Feynman gauge> convention from part <3/a>. Put $s=q^2$, $t=r_t^2$, $u=r_u^2$, the <Mandelstam variables>, and define the three-boson tensor with incoming <four-momenta> $(q,-k_1,-k_2)$:
$$
V^{\lambda\mu\nu}
=g^{\lambda\mu}(q+k_1)^\nu
+g^{\mu\nu}(k_2-k_1)^\lambda
-g^{\nu\lambda}(q+k_2)^\mu.
$$
The contributions, with external <Dirac spinors> included and both external <gauge boson> <propagators> amputated, are
$$
\boxed{\begin{aligned}
\mathcal M_t^{ab\mu\nu}
&=-g^2\bar v(p_2)\,t^bt^a\gamma^\nu
\frac{\not r_t+m}{t-m^2+i0}\gamma^\mu u(p_1),\\
\mathcal M_u^{ab\mu\nu}
&=-g^2\bar v(p_2)\,t^at^b\gamma^\mu
\frac{\not r_u+m}{u-m^2+i0}\gamma^\nu u(p_1),\\
\mathcal M_s^{ab\mu\nu}
&=-ig^2\epsilon^{abc}\bar v(p_2)t^c\gamma_\lambda u(p_1)
\frac{V^{\lambda\mu\nu}}{s+i0},\\
\mathcal M^{ab\mu\nu}&=\mathcal M_t^{ab\mu\nu}
+\mathcal M_u^{ab\mu\nu}+\mathcal M_s^{ab\mu\nu}.
\end{aligned}}
$$
For example, $i\mathcal M_t$ contains $(+ig)^2i=-ig^2$. In the annihilation channel, the current vertex, <propagator>, and three-boson vertex give $(+ig)(-i)g=g^2$, so division by the defining overall $i$ gives $-ig^2$. These relative signs are essential for the cancellation in part <3/d>. The antisymmetric colour factor makes the last expression symmetric under simultaneously exchanging $(a,\mu,k_1)$ and $(b,\nu,k_2)$, as required for identical outgoing <gauge bosons>. The <on shell> external <Dirac spinors> obey $(\not p_1-m)u(p_1)=0$, $\bar v(p_2)(\not p_2+m)=0$.
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