Myhill-Nerode - Prove irregularity for $\{a^{n^3}\}$
More generally, you can use Myhill–Nerode to prove the following characterization:

For $A \subseteq \mathbb{N}$, let $L(A) = \{ a^n : n \in A \}$.
The language $L(A)$ is regular iff $A$ is eventually periodic.

In particular, if $A$ is an infinite set with density zero then $L(A)$ is not regular. Here density zero means:
$$
\lim_{n\to\infty} \frac{|A \cap \{0,\ldots,n-1\}|}{n} = 0.
$$
This shows that $\{a^{n^k} : n \in \mathbb{N}\}$ is not regular for any $k \geq 2$.

Let us prove the claimed characterization. If $A$ is periodic, say with period $m$, then $L(A)$ is clearly regular: it is accepted by a circular DFA with $m$ states. If $A$ is eventually periodic then there is a periodic $B$ such that $L(A),L(B)$ differ in finitely many words, and so standard closure operations show that $L(A)$ is regular.
Now suppose that $L(A)$ is regular. The equivalent class of $a^m$ is the set $E_m = \{ k \in \mathbb{N} : m+k \in A \}$. Since there are finitely many equivalence classes, there exist $m &lt; p$ such that $E_m = E_p$, that is, for all $k \in \mathbb{N}$, $m + k \in A$ iff $p + k \in A$. In other words, for all $\ell \geq m$, $\ell \in A$ iff $\ell + (p-m) \in A$. Hence $A$ is eventually periodic (with period $p-m$).