Your language could be rewritten more clearly as:
$\{ a^m \mid m \text{ is a perfect cube} \}$
Now you'd have to find a word from this language, longer than the pumping length $p$, that can not be pumped. We choose as a candidate: $a^{p^3}$.
We name our three substrings $x, y, z$, which must be of the form:
$x = a^i$
$y = a^j$
$z = a^{(p^3 - i - j)}$
The pumping lemma restricts $j$ to be greater than $0$ and $i + j \le p$. It's sufficient to show that there exists at least one integer $k \ge 0$, such that $xy^kz$ is not a valid word in the language. Because of our separation, we need to find a $k$ such that $p^3 + (k - 1)j$ is not a perfect cube.
Let's say $k = 2$, which gives us $p^3 + j$, strictly increasing with respect to $j$, having a maximum value of $p^3 + p$.
We also know that the smallest perfect cube larger than $p^3$ is $(p + 1)^3 = p^3 + 3p^2 + 3p + 1$.
Because $p^3 \lt (p^3 + p) \lt (p+1)^3, \forall p \gt 0$, our pumped string's length can't be a perfect cube (because it's strictly between two consecutive perfect cubes).
So the language contains at least one word longer than the pumping length that can't be pumped, which means the language isn't regular.
