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.