Why does «Massey cube» of an odd element lie in 3-torsion? The cup product is supercommutative, i.e the supercommutator $[-,-]$ is trivial at the cohomology level — but not at the cochain level, which allows one to produce various cohomology operations.
The simplest (in some sense) of such (integral) operations is the following «Massey cube». Suppose $a$ is an integral $k$-cocycle, $k$ is odd; $[a,a]=0\in H^{2k}$, so $[a,a]=db$ (where $b$ is some cochain); define $\langle a\rangle^3:=[a,b]\in H^{3k-1}$ (clearly this is a cocycle; it doesn’t depend on choice of $b$, since by supercommutativity $[a,b’-b]=0\in H^{3k-1}$ whenever $d(b-b’)=0$).
The question is,

why $\langle a\rangle^3$ lies in 3-torsion?

For $k=3$, for example, this is true since $H^8(K(\mathbb Z,3);\mathbb Z)=\mathbb Z/3$, but surely there should be a more direct proof? (Something like Jacobi identity, maybe?)

Recall that $d(x\cup_1y)=[x,y]\pm dx\cup_1 y\pm x\cup_1dy$.
In particular, in the definition from the question one can take $b=a\cup_1a$. So $\langle a\rangle^3=[a,a\cup_1a]$.
Now $d((a\cup_1a)\cup_1a)=[a,a\cup_1a]+(d(a\cup_1 a))a=\langle a\rangle^3+[a,a]\cup_1 a$. Now by Hirsch formula $a^2\cup_1a=a(a\cup_1a)+(a\cup_1a)a=\langle a\rangle^3$.
So $$3\langle a\rangle^3=d(a\cup_1a\cup_1a)=0\in H^{3k-1}.$$
