By simpler multiples, to invert $\, a - f\,$ where $\,a\,$ is invertible, say $\,\color{#0a0}{ab = 1},$ and $\,f\,$ is nilpotent $\color{#c00}{f^n = 0},\,$ we invert its *simpler multiple* $\, a^n-\color{#c00}{f^n} = \color{#0a0}{a^n},\,$ with obvious inverse $\,\color{#0a0}{b^n},\,$ explicitly

$\ \ \ \ \ \ \ \ \ \color{#0a0}{ab=1},\, \color{#c00}{f^{\large n} = 0}\ \Rightarrow\ \overbrace{\dfrac{1}{a-f} = \dfrac{a^{\large n-1}\!+\!\cdots\! +\! f^{\large n-1}}{\!\!\!\!\!\color{#0a0}{a^{\large n}}-\color{#c00}{f^{\large n}}}}^{\large \text{check via cross multiply}} =\, \color{#0a0}{b^{\large n}}(a^{\large n-1}+\cdots + f^{\large n-1})$
$\!\begin{align}{\rm so}\ \ &\color{#0a0}{3(3)=1},\, \color{#c00}{f^{\large 3} = 0}\ \Rightarrow\ \dfrac{1}{3-f} = \dfrac{3^{\large 2}+\,3f\,+\, f^{\large 2^{\phantom{|^{|^.}}\!\!\!\!\!}}}{\color{#0a0}{3^{\large 3}-\color{#c00}{f^{\large 3}}}}\ \ \, =\, \ \ \color{#0a0}{3^{\large 3}}(3^{\large 2}\! +3f + f^{\large 2}) = \bbox[5px,border:1px solid #c00]{2x+3}\\[.1em]
&{\rm because}\ \ \ \color{#c00}{2^{\large 3}\mid f^{\large 3}}\ \ {\rm by}\ \ 2\mid f = -6x-4x^2,\,\ {\rm to\ invert}\ \ 3\!-\!f = 3\!+\!6x\!+\!4x^2 \in \smash[b]{\Bbb Z_{\large \color{#c00}{2^{\Large 3}}}}\end{align}$

Generally it is easy to prove that a polynomial is a unit iff its constant term is a unit and all other coefficients are nilpotent (the method of proof there can be made constructive - similar to above).

This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses. Analogous methods may be employed for for computing remainders via modular arithmetic.