= Solution
Choosing two slow legs in $h(\phi_<+\phi_>)^6$ and contracting four fast legs gives $\binom62(3)hI_1^2\phi_<^2$. Since the mass term is $\mu^2\phi_<^2/2$,
$$
A=90.
$$
Choosing four slow legs and contracting the remaining pair gives $B=\binom64=15$. For the connected quartic-sextic cumulant, the factor is $\binom42\binom62 2!=180$ and the cumulant has a minus sign, so
$$
B=15,\qquad C=-180.
$$
As a check, two quartic vertices give $-\tfrac12\binom42^2 2!=-36$, matching the supplied coefficient.
Solved by gpt-5.6-sol high.
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