Exercise from Serre's "Trees" - prove that a given group is trivial

Observe first that $$ x_2x_1=x_1^2x_2,\quad x_3x_2=x_2^2x_3,\quad x_1x_3=x_3^2x_1 $$ and more generally $$ x_2^jx_1^k = x_1^{2^jk} x_2^j,\qquad x_3^jx_2^k = x_2^{2^jk} x_3^j,\qquad x_1^jx_3^k = x_3^{2^jk} x_1^j $$ for any $j,k\in\mathbb{N}$. Then $$ x_1^2 x_2^2 x_3 \;=\; x_1^2 x_3 x_2 \;=\; x_3^4 x_1^2 x_2 \;=\; x_3^4 x_2 x_1 \;=\; x_2^{16}x_3^4x_1 \;=\; x_2^{16}x_1x_3^2 \;=\; x_1^{2^{16}}x_2^{16}x_3^2, $$ and solving for $x_3$ gives $$ x_3 \,=\, x_2^{-16}x_1^{2-2^{16}}x_2^2. $$ Then $$ x_2^2 \;=\; x_3x_2x_3^{-1} \;=\; x_2^{-16}x_1^{2-2^{16}}x_2 x_1^{2^{16}-2}x_2^{16} \;=\; x_2^{-16}x_1^{2^{16}-2}x_2^{17} $$ and it follows that $x_2 = x_1^{2^{16}-2}$. In particular, $x_2$ and $x_1$ must commute, so the relation $$ x_2x_1 = x_1^2x_2 $$ proves that $x_1 = 1$, and hence $x_2=1$ and $x_3=1$.

Note: The main trick here was the initial string of equations. In general, we you have relations in a group that pairs of elements "almost" commute, e.g. $x_2x_1=x_1^2 x_2$, you can get a lot of mileage from trying to implement the equations $$ abc \;=\; acb \;=\; cab \;=\; cba \;=\; bca \;=\; bac \;=\; abc. $$ For "real" commutation you get the same $a$, $b$, and $c$ at the end, but for "fake" commutation you usually get something slightly different than what you started with, in this case something with $x_3^2$ instead of $x_3$. Once we found an expression for $x_3$ in terms of $x_1$ and $x_2$ the rest was pretty straightforward.