Big O Notation Simplification in the fraction form
It is indeed confusing if you are expecting some kind of simpler form.

However, ${n^3}/{\log^{3}n}$ is already in the simplest form in term of big-$\Theta$ notation. Consequently, it is also sort of the best form in term of big-$\Omega$ notation or big-$O$ notation. It is certainly true that $n^3/\log^3n = O(n^3)$. However, that kind of big-$O$ bound are usually frowned upon since it is obviously not tight, $n^3/\log^3n = o(n^3)$. In fact, in a similar vein, many time-complexities is described by $\Theta(n\log n)$, $O(n\log n)$ or $\Omega(n\log n)$, where the factor $\log n$ cannot be or is not removed.

The point is that although you can ignore the asymptotically insignificant terms in a sum to arrive at a simpler big-$O$ (or big-$\Theta$, big-$\Omega$) bound, you cannot do away with any unbounded factors without losing significant asymptotical information. For example, it is correct that $n^3+\log^{3}n=O(n^3)$ or $n^3+\log^{3}n=\Omega(n^3)$ or $n^3+\log^{3}n=\Theta(n^3)$, but neither $n^3/\log^{3}n=\Omega(n^3)$ nor $n^3\log^{3}n=O(n^3)$ is correct.

If you want to compare $n^3/\log^3n$ with $(2^{n})^{0.008}$, you can proceed as the following.
$$\text{Since } n^3/\log^3n = O(n^3) \text { and } n^3=o((2^{0.008})^n), \text{ so } n^3/\log^3n =o((2^{0.008})^n)$$
where the second equality comes from the fact $2^{0.008}&gt;1$ and the fact that any polynomial function grows slower than any increasing exponential function, which is proven in another answer of mine. Please note that I am using the little $o$-notation, which means basically "asymptotically growing slower than".

If you want to compare $n^3/\log^3n$ with $2^{n^{0.008}}$, you can proceed as the following.
$$\text{Since } n^3/\log^3n = O(n^3) \text { and } n^3=2^{3\log_2n} = o\left(2^{(n^{0.008})}\right), \text{ so } n^3/\log^3n =o\left(2^{n^{0.008}}\right)$$
where the second equality comes from the fact $\log_2n=o(n^{0.008})$, which is also proven in that answer of mine since $\log_2n=(\log_2e)\log n$.

You can check the reference question and answers on sorting functions for good advices and more detailed and systematic treatment.