Big-theta of $\log^2(n)$ I have been trying to understand weather I can reduce this to more simple possible terms? : $$\Theta(n^3)+\Theta(n^2 \log^2(n))$$ I know that $O(\log^2(n))=O(n^a)$ for $a > 0$. But I'm not sure if that true with theta notation.

The theta notation means a **strict upper bound** , which means that no function more closely bounds the complexity than the function inside the parentheses. For instance, you could say Heapsort is bounded by $O(n\log n)$, $O(n^2)$, or $O(n^n)$ steps, but you could only say that it is bounded by $\Theta(n\log n)$ steps. By the same token, Heapsort is bounded by $\Omega(n)$, $\Omega(n\log(n))$, and $\Omega(1)$. Now, to answer your specific question, since this is asymptomatic, you can add them together and ignore the lower order terms.