Question

n³ - 1 ⋮ n + 1

Answer 1

Ta có: \(n^3-1⋮n+1\)

\(\Leftrightarrow n^3+n^2-n^2-n+n+1-2⋮n+1\)

\(\Leftrightarrow n^2\left(n+1\right)-n\left(n+1\right)+\left(n+1\right)-2⋮n+1\)

\(\Leftrightarrow\left(n+1\right)\left(n^2-n+1\right)-2⋮n+1\)

mà \(\left(n+1\right)\left(n^2-n+1\right)⋮n+1\)

nên \(-2⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(-2\right)\)

\(\Leftrightarrow n+1\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{0;-2;1;-3\right\}\)

Vậy: \(n\in\left\{0;-2;1;-3\right\}\)