When $\log^k(n) = \log (\log (...\log(n)))$,
$$(3 \log^2 n + 55 \log(n^{10}) + 8 \log n) \cdot \log n \\
= \Theta((\log^2(n)+\log(n)+\log(n))\log(n)) \\
= \omega(\log^3(n)).$$
In fact, anything growing strictly slower than $\log(n)\log\log(n)$ will hold inside the $\omega$, so $\log^{10}(n)$ will also fit.
When $\log^k(n)=(\log(n))^k$,
$$(3 \log^2 n + 55 \log(n^{10}) + 8 \log n) \cdot \log n \\
= \Theta(\log^2(n)+\log(n)+\log(n))\cdot \log(n) \\
=\omega(\log(n)).$$
To see the big-$\Theta$ bounds are correct, verify that
$$\lim_{n\rightarrow \infty} \frac{(3 \log \log n + 55 \log(n^{10}) + 8 \log n) \cdot \log n}{(\log \log(n) + \log(n)+\log(n))\log(n)}=O(1).$$
$$\lim_{n\rightarrow \infty}\frac{(3 \log(n)\log(n) + 55 \log(n^{10}) + 8 \log n) \cdot \log n}{(\log(n)\log(n)+\log(n)+\log(n))\cdot \log(n)}=O(1).$$
To see the $\omega$ bounds are correct:
$$\lim_{n\rightarrow \infty}\frac{(3 \log\log n + 55 \log(n^{10}) + 8 \log n) \cdot \log n}{\log\log\log(n)}=\infty.$$
$$\lim_{n\rightarrow \infty}\frac{(3 \log(n)\log(n) + 55 \log(n^{10}) + 8 \log n) \cdot \log n}{\log(n)}=\infty.$$
You can derive the $O$ bounds and $\Omega$ bounds from the $\Theta$ bounds.
I may have misplaced a few terms or two when typesetting; in which case, please point out.