= Solution
{c}
The two <Bhabha scattering> <tree-level Feynman diagrams> are the two panels of <2/b/i/Solution> with a <photon> mediator: $s$-channel annihilation with internal <momentum> $p_1+p_2$, and $t$-channel exchange with internal <momentum> $p_1-p_3$. Each <interaction vertex> preserves the <electron> species. There is no independent $u$-channel <Feynman diagram>, since the outgoing <electron> and <Positron> are distinguishable.
Removing the <color charge> factor from <2/b/i/Solution> and replacing $g$ by $e$ gives
$$
\overline{|\mathcal M_s|^2}=2e^4\frac{t^2+u^2}{s^2}.
$$
<Crossing symmetry> interchanges $s$ and $t$ for the exchange <Feynman diagram>, giving
$$
\overline{|\mathcal M_t|^2}=2e^4\frac{s^2+u^2}{t^2}.
$$
Squaring the sum of the two <scattering amplitudes> also produces the <interference> term $2\operatorname{Re}\overline{\mathcal M_s\mathcal M_t^*}$. The given third term must be this cross term: it has both <propagator> denominators $st$, whereas a squared individual <Feynman diagram> has $s^2$ or $t^2$. With the correct relative <fermionic sign>, its value is $4e^4u^2/(st)$. Thus
$$
\boxed{\overline{|\mathcal M|^2}=2e^4\left[\frac{t^2+u^2}{s^2}+\frac{s^2+u^2}{t^2}+\frac{2u^2}{st}\right].}
$$
The coefficient of the last displayed fraction is positive, but $s>0$ and $t<0$ in the physical scattering region, so this <interference> is destructive. The relative sign originates in the two possible external <fermion> contractions; changing an external-state phase changes both <scattering amplitudes> together.<Quantum Field Theory by David Tong>[§6.6, Electron Positron Scattering: annihilation and exchange spinor contractions and their relative sign]{ref}
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