Problem: Are there integers $a$ and $b$
such that $a^5b+3$ and $ab^5+3$
are both perfect cubes of integers?

Solution: No, there do not exist such $a$ and $b$.

We prove this in two cases.
\begin{itemize}
  \ii Assume $3 \mid ab$.
  WLOG we have $3 \mid a$,
  but then $a^5b+3 \equiv 3 \pmod 9$, contradiction.

  \ii Assume $3 \nmid ab$.
  Then $a^5b+3$ is a cube not divisible by $3$,
  so it is $\pm 1 \bmod 9$,
  and we conclude
  \[ a^5b \in \{5,7\} \pmod 9. \]
  Analogously \[ ab^5 \in \{5,7\} \pmod 9. \]
  We claim however these two equations cannot hold simultaneously.
  Indeed $(ab)^6 \equiv 1 \pmod 9$ by Euler's theorem,
  despite $5 \cdot 5 \equiv 7$, $5 \cdot 7 \equiv 8$,
  $7 \cdot 7 \equiv 4$ mod $9$.
\end{itemize}