Exercise from Serre's "Trees" - prove that a given group is trivial In Serre's book "Trees" on page 10 the following exercise is given:

Show that the group defined by the presentation 
  $$x_2x_1x_2^{-1}=x_1^2, \hspace{7pt} x_3x_2x_3^{-1}=x_2^2, \hspace{7pt} x_1x_3x_1^{-1}=x_3^2$$
  is trivial.

Comparing to what was done before, clearly, the approach he used to prove that a smilarily defined group is infinite - will fail.
My question is - how would you approach this? just sequentailly substiuting one word into another to show that, say, $x_1=1$? Is there any smarter way that just brute-forcing something that cancels out?
Thanks in advance for any help.
N.B: Being an exercise from "Trees", I would expect it to be asked here before. I did my best trying to find it - but  couldn't. So I'm sorry if it turns out to be a duplicate...
 A: 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.
