Proving $n+3 \mid 3n^3-11n+48$ I'm really stuck while I'm trying to prove this statement:
$\forall n \in \mathbb{N},\quad (n+3) \mid  (3n^3-11n+48)$.
I couldn't even how to start.

If you wrote it backwards by accident then the proof is by
$$(3n^2 - 9n + 16)(n+3) = 3n^3 - 11n + 48$$

if you meant what you wrote then $n=0\;$ gives a counter examples since $48$ doesn't divide $3$.
