Show that $1+a^2b$ is invertible in the ring A Let $A$ be a ring and $a,b \in A$ such that $\exists\, n \in \mathbb{N}^\ast$ so that $(aba)^n=0$. Show that $1+a^2b$ is invertible. We have $(aba)^n=0 \implies -(aba)^n=0 \implies 1-(aba)^n=1 $$$\implies (1-aba)(1+aba+(aba)^2+\ldots+(aba)^{n-1})=1,$$ which means that $1-aba$ is invertible. How could that be used(if possible) to prove $1+a^2b \in U(A)$?

What you've really shown is that if $x^n=0$, then $1-x\in U(A)$, which also gives $1+x\in U(A)$ since $(-x)^n=0$, so we really only need to show that $(a^2b)^N=0$ for some $N$, and by a tiny bit of trial and error it seems to me that $N=n+1$ works.

We have

$$(a^2b)^{n+1}=aabaab\cdots aab=a(aba)^nab=0,$$

so $1+a^2b\in U(A)$.