Let $K$ be a non commutative field and $a,b \in K$ such that $ab \neq ba$. Show that : $$ a=\big[b-(a-1)^{-1}b(a-1)\big]\big[a^{-1}ba-(a-1)^{-1}b(a-1)\big]^{-1}. $$
I tried to move the second term of the right hand side to the left and develop but it seem to go nowhere.
Multiply on both sides on the right by $[a^{-1}ba-(a-1)^{-1}b(a-1)]$:
$a[a^{-1}ba-(a-1)^{-1}b(a-1)] = [b-(a-1)^{-1}b(a-1)]$
Distribute the $a$:
$aa^{-1}ba-a(a-1)^{-1}b(a-1) = ba-a(a-1)^{-1}b(a-1) = b-(a-1)^{-1}b(a-1)$
where $aa^{-1}ba=eba=ba$ where $e$ is the identity of multiplication in the field
Group together on both sides of the '$=$' that have common left or right factors:
$ba-b=a(a-1)^{-1}b(a-1)-(a-1)^{-1}b(a-1)$
Factor out the $b$ on the left and $(a-1)^{-1}b(a-1)$ on the right:
$b(a-1)= (a-1)[(a-1)^{-1}b(a-1)]$
then $b(a-1)=b(a-1)$ because $(a-1)(a-1)^{-1}=e$, the multiplicative identity.
