Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$. 
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.

My (attempted) proof:
Suppose that $a^2 \mid b$ and $b^3 \mid c$, where $a$, $b$, and $c$ are integers. 
Therefore, we have that $a^2 k_1 = b$ and $b^3 k_2 = c$ for some $k_1, k_2 \in \mathbb{Z}$.
I'm unsure of how to proceed from here. If I cube $a^2 k_1 = b$ to get $a^6 k_1 = b^3$, and then substitute this into $b^3 \mid c$, then I get $a^6k_1k_2 = c$, which isn't useful. The same happens if I do the other substitution. 
I would greatly appreciate it if people could please take the time to clarify how I should be proceeding.

For some integers $k,l$, we have $a^2k =b, b^3l = c$.
Hence $c^3 = b^9l^3 = a^{18}k^9l^3 = (a^{10}k^5)a^8k^4l^3  = b^5a^4(a^4k^4l^3)$
