Question: > 
> Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
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**My attempt**: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then, 
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$

Not sure how to get rid of $3n^2 + 3n +2$.

Answer: 
Hint:
$$ 3n^2+3n+2<n^3+n<3^n. $$
You proved correctly that $(n+1)^3+(n+1)<3^n+3n^2+3n+2$. So, prove now that$$(\forall n\geqslant4):3n^2+3n+2\leqslant2\times3^n.$$This is easy to do using induction again.