= Solution
{c}
For the outgoing massless <gauge bosons>, take $k_1^2=k_2^2=0$. Work away from internal poles, with the <Feynman i-epsilon prescription> understood. The <Dirac equations> and <four-momentum conservation> give
$$
\begin{aligned}
\bar v\not k_2&=\bar v(\not r_t-m),\\
\not k_2u&=-(\not r_u-m)u.
\end{aligned}
$$
Multiplying by the neighboring <Dirac propagator> numerator cancels its denominator. The two fermion-exchange diagrams therefore yield the <matrix commutator>
$$
\begin{aligned}
k_{2\nu}(\mathcal M_t+\mathcal M_u)^{ab\mu\nu}
&=g^2\bar v\gamma^\mu(t^at^b-t^bt^a)u\\
&=ig^2\epsilon^{abc}J_c^\mu,\qquad
J_c^\lambda=\bar v(p_2)\gamma^\lambda t^cu(p_1).
\end{aligned}
$$
The annihilation <vector current> obeys $q_\lambda J_c^\lambda=\bar v(\not p_1+\not p_2)t^cu=0$. Direct contraction of the three-boson tensor gives
$$
k_{2\nu}V^{\lambda\mu\nu}
=s g^{\lambda\mu}-q^\lambda k_2^\mu-k_2^\lambda k_1^\mu.
$$
Thus
$$
k_{2\nu}\mathcal M_s^{ab\mu\nu}
=-ig^2\epsilon^{abc}\left(J_c^\mu-\frac{J_c\cdot k_2}{s}k_1^\mu\right).
$$
The terms proportional to $J_c^\mu$ cancel, leaving the <non-Abelian Ward identity for fermion annihilation>
$$
\boxed{k_{2\nu}\mathcal M^{ab\mu\nu}
=ig^2\epsilon^{abc}\frac{J_c\cdot k_2}{s}\,k_1^\mu.}
$$
The open-index tensor is generally nonzero: neither $\epsilon^{abc}$ nor $J_c\cdot k_2$ is constrained to vanish. For a concrete nonzero configuration, take <centre-of-momentum frame> kinematics $p_1=(E,0,0,p)$, $p_2=(E,0,0,-p)$, $p=\sqrt{E^2-m^2}$, $k_1=(E,E,0,0)$, $k_2=(E,-E,0,0)$. Standard <Dirac spinors> with two-component spin labels $\chi=(1,0)^T$ for $u$ and $\eta=(0,1)^T$ for $v$ give the spinor current $(0,2E,2iE,0)$ and its contraction with $k_2$ is $2E^2$. Choosing both colour spinors $(1,0)^T$ and $(a,b,c)=(1,2,3)$ multiplies this by $1/2$, still nonzero. This explicitly rules out an unconditional tensor identity $k_{2\nu}\mathcal M^{ab\mu\nu}=0$.
For a physical <polarization vector> $\varepsilon_{1\mu}^*$, however, $k_1\cdot\varepsilon_1^*=0$, and therefore
$$
\boxed{\varepsilon_{1\mu}^*\,k_{2\nu}\mathcal M^{ab\mu\nu}=0.}
$$
This is the required decoupling of a pure-gauge <polarization vector> on leg 2 when the other leg is physical. Interchanging the legs gives the corresponding identity for leg 1. The role of the three-boson diagram is explicit: it cancels the noncommuting colour-current term left by the two fermion exchanges.
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