$A+A^2B+B=0$ implies $A^2+I$ invertible? Let $A$ and $B$ be two square matrices over a field such that $A+A^2B+B=0$.
Is it true that $A^2+I$ is always invertible ?

${\bf Hint}\  \ 1\!+\!aa\mid \color{#c00}a\,\Rightarrow\, 1\!+\!aa \mid 1 = 1\!+\!aa-\color{#c00}aa,\ $ e.g. applying this to the special case at hand:
$\quad\ \ \color{#0a0}{{-}(1\!+\!aa)b} =  \color{#c00}{a},\ $ hence $\ 1 = 1\!+\!aa-\color{#c00}{a}a = 1\!+\!aa\color{#0a0}{+(1\!+\!aa)b}a=(1\!+\!aa)(1\!+\!ba) $
Remark $\ $ Essentially we exploit the universality of the Bezout identity $\,1\!+\!aa-\color{#c00}aa = 1.$
