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.

Let's try to find the $n$ such that the simpler inequality
$9 n^3 < 8^n$ holds. If this is true, the desired inequality is true.
Computation will settle the smaller values.
Taking the cube root, this is the same as
$n 9^{1/3} < 2^n$. Since $9^{1/3} < 3$, this is true when
$3n < 2^n$.
This is true when $n=4$ ($12 < 16$). 
If is true for $n \ge 4$,
$3(n+1) = 3n(1+1/n) < 2^n(1+1/4) \le 2^{n+1}$.
The result is thus true for $n \ge 4$. 
For $n=3$, $9 n^3-4= 239$ and $8^3 = 512$, so it is true.
For $n=2$, $9 n^3-4 = 68$ and $8^2=64$ so it is false.
