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 $f(n)=n^3-3$, and let $g(n)=8^n$. We compute a little, to see what is going on.
We have $f(0) \le g(0)$; $f(1)\le g(1)$; $f(2) > g(2)$; $f(3) \le g(3)$; $f(4) \le g(4)$. Indeed $f(4)=573$ and $g(4)=4096$, so it's not even close. 
The exponential function $8^x$ ultimately grows incomparably faster than the polynomial $9x^3-3$. So it is reasonable to conjecture that $9n^3-3 \le 8^n$ for every non-negative integer $n$ except $2$. 
We will show by induction that $9n^3-3 \le 8^n$ for all $n \ge 3$. It is natural to work with ratios. We show that 
$$\frac{8^n}{9n^3-3} \ge 1$$ 
for all $n \ge 3$. The result certainly holds at $n=3$. 
Suppose that for a given $n \ge 3$, we have $\frac{8^n}{9n^3-3} \ge 1$. We will show that $\frac{8^{n+1}}{9(n+1)^3-3} \ge 1$.
Note that
$$\frac{8^{n+1}}{9(n+1)^3-3}=8 \frac{9n^3-3}{9(n+1)^3-3}\frac{8^n}{9n^3-3}.$$
By the induction hypothesis, we have $\frac{8^n}{9n^3-3} \ge 1$.  So all we need to do is to show that 
$$8 \frac{9n^3-3}{9(n+1)^3-3} \ge 1,$$
or equivalently that 
$$\frac{9(n+1)^3-3}{9n^3-3} \le 8.$$ 
If $n\ge 3$, the denominator is greater than $8n^3$, and the numerator is less than $9(n+1)^3$.  Thus, if $n \ge 3$, then 
$$\frac{9(n+1)^3-3}{9n^3-3} <\frac{9}{8}\frac{(n+1)^3}{n^3}=\frac{9}{8}\left(1+\frac{1}{n}\right)^3.$$
But if $n \ge 3$, then $(1+1/n)^3\le (1+1/3)^3<2.5$, so $\frac{9}{8}(1+1/n)^3<8$, with lots of room to spare. 
