Given a torsion-free hyperbolic group $G$, does there exist a number $n(G)$ such that for any $x,y,z\in G$, $x^n y^n z^n =1$ implies that $x$, $y$, and $z$ commute pairwise?


Some musings/questions...
When $G$ is free, the result is true for $n=2$. 


Clearly, this does not generalize to hyperbolic groups (e.g. non-orientable surface groups of genus 3). 


Somewhat related: given any 3 non-commuting elements $x,y,z\in G$ there is a number $n(x,y,z)$ such that $\langle x^n, y^n , z^n \rangle $ is free. Is this true??..definitely in case of two elements (Gromov). 



Lyndon-Schützenberger for torsion-free hyperbolic groups
The second statement is true if $x,y,z$ are not torsion as proved by Arzhantseva and, independently, by Kapovich and Weidmann (see Arzhantseva, Goulnara N.
A dichotomy for finitely generated subgroups of word hyperbolic groups. Topological and asymptotic aspects of group theory, 1–10, 
Contemp. Math., 394, Amer. Math. Soc., Providence, RI, 2006 and Kapovich, Ilya; Weidmann, Richard
Nielsen methods and groups acting on hyperbolic spaces. 
Geom. Dedicata 98 (2003), 95–121. This also implies the first statement (about the equation $x^ny^nz^n=1$) in the torsion-free case.


