Suppose $F_n$ is a free group of rank $n$. It is a rather well known fact, that $b_3(n) = [F_n : V_{\{x^3\}}(F_n)]$ is finite for all $n \in \mathbb{N}$. Is there a some sort of formula for $b_3(n)$? Here $V_Q$ is the verbal subgroup for the collection of group words $Q$.
The solution of a similar problem for $b_2(n) = [F_n : V_{\{x^2\}}(F_n)]$ is quite obvious: $b_2(n) = 2^n$ (as $C_2^n$ is the only $n$-generated group of exponent $2$) However, similar considerations do not work for $b_3$ (as, for example $C_3 \times C_3$ and $UT(3, 3)$ are both $2$-generated groups with exponent $3$, but have different orders).
However, using that method a lower bound can be constructed: $$b_3(n) = [F_n : V_{\{x^3\}}(F_n)] \geq [F_n : V_{\{x^3, [x, y]\}}(F_n)] = 3^n$$
Attempting to find an upper bound on $b_3(n)$ by counting cube-free words (something similar can also be done to $b_2(n)$) is doomed to fail too, as even for $n = 2$, there are infinitely many of them: For what $n$ is $W_n$ finite?
