Prove for which $n \in \mathbb{N}$: $9n^3 - 3 ≤ 8^n$ A homework assignment requires me to find out and prove using induction for which $n ≥ 0$ $9n^3 - 3 ≤ 8^n$ and I have conducted multiple approaches and consulted multiple people and other resources with limited success. I appreciate any hint you can give me.
Thanks in advance.

Clearly, if $3n\le 2^n$ then $9n^3-3\le (3n)^3 \le (2^n)^3 = 8^n$.
So let us have a look at the (hopefully simpler) problem when $3n \le 2^n$. If we are able to solve this problem, only finitely many cases will remain to be checked.
Claim: $3n\le 2^n$  holds for every $n\in\mathbb N$, $n\ge 4$.
Proof by induction. $1^\circ$  For $n=4$ we have $3n = 12 \le 16 = 2^n$.
$2^\circ$. Suppose the claim is true for $n$, we will verify it for $n+1$.
$3(n+1)=3n\cdot\frac{n+1}n \le 2^n\cdot 2 = 2^{n+1}$.
We have used $3n\le 2^n$ (inductive hypothesis) and $\frac{n+1}n=1+\frac1n\le2$.
