(edited a bit to cover a few questions about the notation)
A slight simplification of David Speyer's argument: his argument using Groebner bases explains that is we degenerate the relation into $a_1a_2a_3=0$, the resulting algebra has the same Hilbert series. Now, the latter algebra $B$ has a very economic resolution of the trivial module by free right modules: $$ 0\to span(a_1a_2a_3)\otimes B\to span(a_1,a_2,a_3)\otimes span_k(a_1a_2a_3)\otimes_k B\to span_k(a_1,a_2,a_3)\otimes_kB\to B\to k\to 0 $$ (the leftmost differential maps $a_1a_2a_3\otimes 1$ to $a_1\otimes a_2a_3$, the next one maps $a_i\otimes 1$ to $a_i$). Computing the Euler characteristics, we get H_B(t)(1-3t+t^3)=1$.
