\(\frac{n^3-n^2-3}{n+1}=\frac{n^2\left(n+1\right)-2n\left(n+1\right)+2\left(n+1\right)-5}{n+1}=n^2-2n+2-\frac{5}{n+1}\)
Để số trên là số nguyên thì 5⋮(n+1)
\(\Rightarrow n+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow n\in\left\{-6;-2;0;4\right\}\)