Finding inverse in non-commutative ring Let $a,b,c,x$ be elements of a unital non-commutative ring. Assume $c$ is an inverse of $1-ab$: $$ c(1-ab) = 1$$
How can I find an inverse for $1-ba$? What I tried: 
Denote the unknown by $x$. Then $x (1-ba) = 1 = x - xba$. I tried to replace $1$ with $c(1-ab)$ but it didn't help because I cannot solve for $x$. I also tried to subtract $x (1-ba) $ from $ c(1-ab)$ but couldn't solve for $x$ either. Any suggestions?
 A: Assuming completeness w.r.t. $(a)$ or $(b)$ or just as a heuristic:
$x = (1 - ba)^{-1} = 1 + ba + baba + \cdots = 1 + b(1 + ab + abab + \cdots)a = 1 + bca$
A: Let's think in term of symbols, since we want $x$ to have a "simple" form.
Suppose that $x(1-ba)=1$. If $x$ could be writen only with symbols $a,b,c$, the result $x(1-ba)$ would also be written in symbols $a,b,c$, so it is natural to search for $x$ of the form $x=1+y$. Also, $y$ must be chosen so that it "cancels" the symbols $ba$.
Notice that the formula $c(1-ab)$ can be rewritten as $c-cab=1$, that is, $cab=c-1$. With that equality, we can get a term with $b$ and rewrite it as a difference of terms without $b$.
Notice that $(1+y)(1-ba)=(1-ba)+y+yba$. In order to rewrite $yba$ without the $b$, we can search for $y$ of the form $y=zca$ for some $z$, so $yba=z(cba)a=z(c-1)a$. Finally, we want that the following holds:
$$(1+zca)(1-ba)=1$$
that is, $1=1-ba+zca-z(cab)a=1-ba+zca-z(c-1)a=1-ba+zca-zca+za=1-ba+za$, which is valid for $z=b$.
Therefore, the natural candidate for the inverse of $1-ba$ is $x=1+bca$, and you can easily show that this is, in fact its inverse.
A: Hint $ $ The inverse can be slickly derived  via geometric power series as below (straightforward algebra then proves that it is indeed an inverse).
$$\begin{eqnarray} \rm (1-ba)^{-1} &=&\rm 1+ \color{#0a0}b\color{#c00}a + \color{#0a0}b\color{c00}{ab}\color{#c00}a + \color{#0a0}b\color{0a0}{abab}\color{#c00}a +\,\cdots\\
&=&\rm 1+ \color{#0a0}b\:\! (1\:\! +\, \color{c00}{ab}\ \ +\ \ \color{0a0}{abab}\,\ +\,\cdots\,)\:\!\color{#c00}a\\
&=&\rm 1+ \color{#0a0}b\:\! (1\,-\,ab)^{-1}\color{#c00}a\end{eqnarray}\qquad\qquad$$
Halmos asks why this works in a famous expository article (excerpted below). When I was a student I became interested in this. It turns out that one can give good (and rigorous) explanations. It can be proved that all such rational identities are essentially consequences of geometric power series expansions. For references see this Mathoverflow question. See also Paul Cohn, A remark on the quasi-inverse of a product, Illinois J. Math, 2003. Cohn wrote this paper in reply to my question regarding his viewpoint on this topic. 


Geometric series. In a not necessarily commutative ring with
  unit (e.g., in the set of all $3 \times 3$ square matrices with real
  entries), if $\,1 - ab\,$ is invertible, then $\,1 - ba\,$ is invertible. However
  plausible this may seem, few people can see their way
  to a proof immediately; the most revealing approach belongs
  to a different and distant subject.
Every student knows that
  $\,1 - x^2 = (1 + x) (1 - x),\,$
  and some even know that
  $\,1 - x^3 =(1+x +x^2) (1 - x).\,$
  The generalization
  $\,1 - x^{n+1} = (1 + x + \cdots + x^n) (1 - x)\,$
  is not far away. Divide by $\,1 - x\,$ and let $\,n\,$ tend to infinity;
  if $\,|x| < 1,\,$ then $\,x^{n+1}$ tends to $\,0,\,$ and the conclusion is
  that
  $\frac{1}{1 - x} = 1 + x + x^2 + \cdots.\,$
  This simple classical argument begins with easy algebra,
  but the meat of the matter is analysis: numbers, absolute
  values, inequalities, and convergence are needed not only
  for the proof but even for the final equation to make
  sense.
In the general ring theory question there are no numbers,
  no absolute values, no inequalities, and no limits -
  those concepts are totally inappropriate and cannot be
  brought to bear. Nevertheless an impressive-sounding
  classical phrase, "the principle of permanence of functional
  form", comes to the rescue and yields an analytically
  inspired proof in pure algebra. The idea is to pretend
  that $\,\frac{1}{1 - ba}\,$ can be expanded in a geometric series (which
  is utter nonsense), so that
  $\,(1 - ba)^{-1} = 1 + ba + baba + bababa + \cdots\,$
  It follows (it doesn't really, but it's fun to keep pretending) that
  $\,(1 - ba)^{-1} = 1 + b (1 + ab + abab + ababab + \cdots) a.\,$
  and, after one more application of the geometric series
  pretense, this yields
  $\,(1 -ba)^{-1} = 1 + b (1 - ab)^{-1} a.\,$
Now stop the pretense and verify that, despite its unlawful
  derivation, the formula works. If, that is, $\, c = (1 - ab)^{-1},\,$ 
  so that $\,(1 - ab)c = c(1 - ab) = 1,\,$ then $\,1 + bca\,$ is the inverse
  of $\,1 - ba.\,$ Once the statement is put this way, its
  proof becomes a matter of (perfectly legal) mechanical
  computation.
Why does it all this work? What goes on here? Why
  does it seem that the formula for the sum of an infinite
  geometric series is true even for an abstract ring in which
  convergence is meaningless? What general truth does
  the formula embody? I don't know the answer, but I
  note that the formula is applicable in other situations
  where it ought not to be, and I wonder whether it deserves
  to be called one of the (computational) elements
  of mathematics. -- P. R. Halmos [1]

[1] Halmos, P.R. $ $ Does mathematics have elements?
Math. Intelligencer 3 (1980/81), no. 4, 147-153  
