I know the usual trick: let $c=(1-ab)^{-1}$ and notice that:
$$(1-ba)(bca) = bca - babca = b(c-abc)a=bc(1-ab)a=ba$$
Etc.
I was wondering how to discover this fact from scratch, without invoking this magical observation. Any help in that?
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11A slick way is to use formal power series as below - see this MO question for why it works. $$\begin{eqnarray} \rm (1-ab)^{-1} &=&\rm 1+ ab + a\color{#c00}{ba}b + a\color{#0a0}{baba}b +,\cdots\ &=&\rm 1+ a (1, +, \color{#c00}{ba}\ \ +\ \ \color{#0a0}{baba},\ +,\cdots)b\ &=&\rm 1+ a (1,-,ba)^{-1}b\end{eqnarray}\qquad\qquad$$ Unless you elaborate, it seems your question is a dupe. – Bill Dubuque Jun 29 '17 at 22:54
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2What's magical about this observation? It's not obvious, but it is elementary, in the sense that it doesn't invoke any big heavy-duty theorems. – Clive Newstead Jun 29 '17 at 22:55
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1@Bill thanks a lot! I don't care what my question seems now, I got what I want. – Cauchy Jun 29 '17 at 23:02