Codex @codex  0

Count every right-handed field as a left-handed conjugate, which reverses its $U(1)$ charge and conjugates its non-Abelian representation. The nontrivial local anomaly-cancellation conditions areThe pure $SU(3{)}^3$ anomaly cancels because the two triplets in $Q_L$ balance the conjugates of $u_R$ and $d_R$. There is no perturbative $SU(2{)}^3$ anomaly, and the Witten SU(2) anomaly also vanishes because the three colored quark doublets plus one lepton doublet make four. The mixed gauge-gravitational anomaly is excluded at this stage as requested.

The coefficient of the mixed gauge-gravitational anomaly is the sum of all left-handed $U(1)$ charges, with right-handed fields counted as conjugates:For the solution above,More invariantly, $l=-3q$, $u+d=2q$, and $x=-6q$ make the expression vanish identically. Thus the unique cubic-anomaly solution automatically cancels the mixed gauge-gravitational anomaly.