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Suppose $G$ is a pro-$p$-group, $p$ odd, and $\mathbb{F}_p$ is given the trivial $G$-action. By skew-symmetry of the cup-product in degree 1, given $\chi\in H^1(G,\mathbb{F}_p)$, we have $\chi\cup\chi=0\in H^2(G,\mathbb{F}_p)$. In fact, in this case, it's even possible to explicitly write $\chi\cup\chi$ as a coboundary -- $\chi\cup\chi=d\left(\binom{\chi}{2}\right)$, the coboundary of "$\chi$ choose 2".
In any case, my question is whether or not there anyone has seen any other tricks of this sort, i.e., for the explicit realization of a trivial cup product as a coboundary. In my specific case, I know a particular cup product is zero since I can force it, via the $G$-equivariance of the cup-product, to land in a known-to-be-trivial eigenspace of $H^2$. I was hoping there was some "eigenspace-averaging" trick similar to the construction of orthogonal idempotents to get my hands on an explicit pre-image, but really, I'd just like to be aware of any tricks for doing this.