= Solution
{c}
Both $s$-channel annihilation and $t$-channel <gluon> exchange are now allowed. Their <spin sum> factors are the <Bhabha scattering> factors with $e\mapsto g$. Their <color charge> tensors, with the external indices of <2/b/i/Solution>, are
$$
c_s=(T^a)_{ji}(T^a)_{kl},\qquad
c_t=(T^a)_{ki}(T^a)_{jl}.
$$
Each squared term has $K_{ss}=K_{tt}=(N_c^2-1)/(4N_c^2)$. The cross term instead has the <color average>
$$
K_{st}=\frac1{N_c^2}\sum_{a,b}\operatorname{tr}(T^aT^bT^aT^b).
$$
To evaluate it, use the <SU(N) fundamental completeness relation>
$$
\sum_a(T^a)_{ij}(T^a)_{kl}=\frac12\left(\delta_{il}\delta_{jk}-\frac1{N_c}\delta_{ij}\delta_{kl}\right).
$$
This follows by completing the orthogonal basis of traceless <Hermitian matrices> with the identity. Since $\operatorname{tr}T^b=0$, it gives $\sum_aT^aT^bT^a=-T^b/(2N_c)$, and hence
$$
\boxed{K_{st}=-\frac{N_c^2-1}{4N_c^3}=-\frac1{N_c}K_{ss}.}
$$
Multiplying the two squared <Bhabha scattering> terms by $K_{ss}$ and its cross term by $K_{st}$ gives
$$
\boxed{\overline{|\mathcal M|^2}=g^4\frac{N_c^2-1}{2N_c^2}
\left[\frac{t^2+u^2}{s^2}+\frac{s^2+u^2}{t^2}-\frac2{N_c}\frac{u^2}{st}\right].}
$$
For the <SU(3)> <gauge group>, this is
$$
\boxed{\overline{|\mathcal M|^2}=\frac{4g^4}{9}\left[\frac{t^2+u^2}{s^2}+\frac{s^2+u^2}{t^2}\right]
-\frac{8g^4}{27}\frac{u^2}{st}.}
$$
The negative <color charge> overlap reverses the sign of the <QED> <interference>. Since $st<0$, this cross contribution is positive in the physical region. No additional <spinor> calculation is needed; all the new information is in the <SU(N) fundamental completeness relation>.
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