I need to prove that the following language is not regular by showing there are infinite pairwise distinct equivalence classes: $$ L = \{a^{n^3} \mid n \geq 1\} \subseteq \{a\}^* $$
Looking at a few examples, I intuitively noticed that for each $i \geq 1$ there is an equivalence class $[a^{i^3}] \neq [a^{j^3}]$ for $i \neq j$ because the number of $a$'s we need to add increases exponentially with $i/j$. Thus for every $i$ we can add a suffix of $a$'s to reach the next higher power $(i+1)^3$ but the length of that suffix is different for every other power, so that each equivalence class only actually contains the representative itself.
Example: $$ \begin{align*} &n = 1 \implies a \\ &n = 2 \implies a^8 \\ &n = 3 \implies a^{27} \\ \vdots \end{align*} $$ Then $$ \begin{align*} & [a] = \{x \in \{a\}^* \mid a \; R_L \; x\} = a \\ & [a^8] = \{x \in \{a\}^* \mid a^8 \; R_L \; x\} = a^8 \\ \vdots \end{align*} $$ Consider $[a]$. For $z = a^7$ we get $az = aa^7 = a^8 \in L$, but starting from $i \geq 2$, we have that $a^{i^3}z \notin L$.
How can I write this down in a more formal/rigorous way?