Proof by Induction $3^n > n^3$ I am trying to prove the following, however I'm stuck at the Induction hypothesis

Prove by induction that, for all integers $n$, if $n\geq 5$, then $3^n>n^3$

What I have Done:
Base Case:
$n = 5$
$3^5 > 5^3$
$243 > 125$ so TRUE
Assume True for $n = k$
$3^k > k^3$ must be TRUE
Should be TRUE for $n = k+1$
$3^k + 3^{k+1} > k^3 + 3^{k+1}$
Im a little stuck here. Any help would be appreciated!
 A: Supposing that $3^k\gt k^3$, you need to prove $3^{k+1}\gt (k+1)^3$.
For inductive step : $$\begin{align}3^{k+1}&\gt 3k^3\\&=k^3+k^3+k^3\\&\gt k^3+3k^2+3^2k\\&=k^3+3k^2+3k+6k\\&\gt k^3+3k^2+3k+1\\&=(k+1)^3.\end{align}$$
A: $3^{k+1}=3*3^k\geq 3k^3$ and it's easy to check that $(k+1)^3\leq 3k^3$ since $3k+1\leq 4k \leq k^3$ and $3k^2\leq k^3$ (because $k\geq 5$).
A: By induction, you assume that for $ n = k,
3^k > k^3$ must be true. We want to show $ 3^{k+1} > (k+1)^3$.
Thus,
$3^{k+1} = 3^k\cdot 3 >3\cdot k^3$, by your induction hypothesis.
So we want to show $ 3^{k+1} > (k+1)^3$, but have $3^{k+1} > 3k^3$.
So just show $3k^3 > (k+1)^3$ for $k>5$. We have
$(k+1)^3 = k^3+3k^2+3k +1 < k^3+k^3+k^3 = 3k^3$.
Therefore,$3^{k+1} > 3k^3 > (k+1)^3 \Longrightarrow 3^{k+1} > (k+1)^3,$ 
and we are done.
A: Your third equations is wrong, I think you are confused with summation induction proofs, you need to prove:
$$3^{k+1}>(k+1)^3$$
You don't need to prove (anyways it will hold true either ways):
$$3^k+3^{k+1}>k^3+k^{3+1}$$
