Let $q>p>10$ be twin primes, i.e., $q=p+2$. Show that every group of order $q^3p^3$ is solvable.
This should be proven without using Burnside's theorem. Looking at the Sylow $p$-subgroup and Sylow $q$-subgoup, their number is $k_p \in {1, q, q^2, q^3}$ and similarly for $k_q$. But I don't see how it helps, how do I countinue from here?
I assume that you know that $p$-groups are solvable ($p$ prime). Now in your case let $P \in {\rm Syl}_p(G)$, then
$$n_p(G)=\#{\rm Syl}_p(G) \in \{1,p+2, (p+2)^2, (p+2)^3\}.$$
Since $n_p(G) \equiv 1$ mod $p$, it follows that $P$ is normal (the first option) or (all mod $p$) $2 \equiv 1$, $4 \equiv 1$ or $8 \equiv 1$. Hence $P \unlhd G$ or $p$ divides one of $\{1, 3, 7\}$. Since $p \gt 10$ is a prime, it must be at least $11$ and all the last possibilities are refuted. Hence $P$ is normal. But then $G/P$ is a $q$-group and hence solvable and $P$ is of course solvable. It follows that $G$ itself is solvable.
This is the proof Derek and Arturo had in mind!