After factorising $a_1$ and $a_2$ into prime numbers, $a_1 = 2^3 \cdot 3$ and $a_2 = 3^4 \cdot 5^2$, we see that we can write $a_n = 2^{\alpha_n} \cdot 3^{\beta_n} \cdot 5^{\gamma_n}$, where for all $n \ge 1$
$$\begin{align*} \alpha_1 &= 3 & \alpha_2 &= 0 & \alpha_{n+2} &= \alpha_{n+1} + 4\alpha_n \\ \beta_1 &= 1 & \beta_2 &= 4 & \beta_{n+2} &= \beta_{n+1} + 4\beta_n + 1 \\ \gamma_1 &= 0 & \gamma_2 &= 2 & \gamma_{n+2} &= \gamma_{n+1} + 4\gamma_n. \end{align*}$$

The number $a_n$ is a perfect cube exactly when each of $\alpha_n, \beta_n$ and $\gamma_n$ is divisible by 3. Therefore, we use the given recursion and calculate the three sequences modulo 3:

We see that the columns repeat in a cycle of length eight and that the values of $n$ for which all three, $\alpha_n$, $\beta_n$ and $\gamma_n$, are divisible by 3, are those of the form $n = 8k + 5$.