Here's an idea. It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis. Now, here are two things which I think are true but which I don't know how to prove:

• The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is trivial.
• $\text{SO}(3)$ is equal to its own commutator subgroup.

If both of these things are true, it follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i \in G_{n-1}$ but $w \not \in G_n$ for some $n$ and setting the $w_i$ to be rotations about a common axis.

Here's an idea. It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis. Now, here are three things which I think are true but which I don't know how to prove:

• The commutator subgroup of a free group can be freely generated by commutators.
• The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is trivial.
• $\text{SO}(3)$ is equal to its own commutator subgroup.

If all of these things are true, it follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i$ are commutators of elements in $G_{n-1}$ but $w \not \in G_{n+1}$ for some $n$ and setting the $w_i$ to be rotations about a common axis. (The first assumption is the one in which I have the least confidence...)

Here's an idea. It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis. Now, here are two things which I think are true but which I don't know how to prove:

• The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is empty.
• $\text{SO}(3)$ is equal to its own commutator subgroup.

It follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i \in G_{n-1}$ but $w \not \in G_n$ for some $n$ and setting the $w_i$ to be rotations about a common axis.

Here's an idea. It's easy to see that there is a pair of rotations not satisfying a word $w$ which does not lie in the commutator subgroup $[F_2, F_2]$ of $F_2$, just by picking two rotations about a common axis. Now, here are two things which I think are true but which I don't know how to prove:

• The intersection $\bigcap G_n$ of the derived series $G_n = [G_{n-1}, G_{n-1}]$ of $F_2$ (where $G_0 = F_2$) is trivial.
• $\text{SO}(3)$ is equal to its own commutator subgroup.

If both of these things are true, it follows that the above argument applies to any $w \in F_2$, by writing $w$ as a word $w_1 ... w_k$ where $w_i \in G_{n-1}$ but $w \not \in G_n$ for some $n$ and setting the $w_i$ to be rotations about a common axis.