Solving the equation $abc=cba$ in the free group. It is known that if two words $a,b$ commute in the free group $F$, then they are powers of the same word, i.e. $a=c^r$ and $b=c^s$, where $c\in F$ and $r,s \in \mathbb Z$.
What happens if there are three words $a,b,c \in F$ such that $a b c=c b a$?
Is there a similar property as above? Or, if not, is there anything that follows from the equation, any information about $a,b$ or $c$? And why?
My goal is to find all solutions for the equation $a b c=c b a$ in the free group of rank two.
 A: Your identity is equivalent to
$$
a b = c (b a) c^ {-1}.
$$
If you choose $a, b$ arbitrarily, you are thus looking for all $c$ that conjugate $b a$ to $a b$. One of the solutions will be $c = a$. All the solutions will thus be of the form $c = a z$, where $z (b a) = (b a) z$, which gets you back to your first statement. 
A: Another way to solve this equation is to replace the product $bc$ with a new variable, $g$ say, and replace $c^{-1}a$ another new variable, $h$ say*. Then your equation becomes
$$
\begin{align*}
abc&=cba\\
c^{-1}abc&=bcc^{-1}a\\
hg&=gh
\end{align*}
$$
Therefore, solutions over the free group $F$ are assignments $(g, h)\rightarrow (w^i, w^j)$ with $i, j\in\mathbb{Z}$ and $w\in F$ such that there is no element $u\in F$ such that $u^k=w$, $k>1$. Substituting in our replacements of $g\leftrightarrow bc$ and $h\leftrightarrow c^{-1}a$, we have all solutions are of the form $$(a, b, c)\rightarrow (vw^i, w^jv^{-1}, v)$$ where $w$ is as above and $v\in F$ is arbitrary.
*This second assignment isn't initially obvious.
