Ta có: \(n^3+n^2+1⋮n+1\)
\(\Leftrightarrow n^2\left(n+1\right)+1⋮n+1\)
mà \(n^2\cdot\left(n+1\right)⋮n+1\)
nên \(1⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(1\right)\)
\(\Leftrightarrow n+1\in\left\{1;-1\right\}\)
hay \(n\in\left\{0;-2\right\}\)
Vậy: Để \(n^3+n^2+1⋮n+1\) thì \(n\in\left\{0;-2\right\}\)