<s>[INST] A Particular Two-Variable System in a Group Suppose $a$ and $b$ are elements of a group $G$.
If $a^{-1}b^{2}a=b^{3}$ and $b^{-1}a^{2}b=a^{3}$, prove $a=e=b$. 
I've been trying to prove but still inconclusive. Please prove to me. 
Thanks very much for proof.
 [/INST] In this answer I give credit to Jyrki Lahtonen for the answer he posted.
There are holes in his post, so I sensed the need for a step by step answer (firstly to convince myself, but also other people in doubt), and so here it is.
$\bbox[5px,border:2px solid]{\begin{array}{cc}a^3=b^{-1}a^2b&(\alpha)\\b^3=a^{-1}b^2a &(\beta)\end{array}}$
$b^{6}=b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)=a^{-1}b^4a\quad(\alpha)$
$b^{12}=b^6b^6=(a^{-1}b^4a)(a^{-1}b^4a)=a^{-1}b^8a\quad(\alpha)$
$b^{18}=b^{12}b^6=(a^{-1}b^8a)(a^{-1}b^4a)=a^{-1}b^{12}a\quad(\alpha)$
$b^{27}=b^{18}b^6b^3=(a^{-1}b^{12}a)(a^{-1}b^4a)(a^{-1}b^2a)=a^{-1}b^{18}a\quad(\alpha)$

$b^{27}=a^{-1}b^{18}a=a^{-2}b^{12}a^2=a^{-3}b^8a^3=\ (\beta)\ =(b^{-1}a^2b)^{-1}b^8(b^{-1}a^2b)=(b^{-1}a^{-2}b)b^8(b^{-1}b^{-1}a^2b)=b^{-1}a^{-2}b^8a^2b=b^{-1}(a^{-1}(a^{-1}b^8a)a)b=b^{-1}(b^{18})b=b^{18}$

$b^{27}=b^{18}\Rightarrow b^9=1$
$b^9=b^3b^3b^3=(a^{-1}b^2a)(a^{-1}b^2a)(a^{-1}b^2a)=(a^{-1}b^6a)=1\Rightarrow b^6=1\quad (\beta)$
$b^3=b^9b^{-6}=1=a^{-1}b^2a\Rightarrow b^2=1$
$b=b^3b^{-2}=1$ and then $a^3=b^{-1}a^2b=a^2\Rightarrow a=1\quad (\alpha)$.
$\bbox[5px,border:2px solid]{a=b=1}$
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