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.

I keep your notations and write $a^2 k_1 = b$ and $b^3 k_2 = c$. Then we have 
$$c^3=b^9k_2^3=b^2b^7k_2^3=(a^4k_1^2)b^7k_2^3$$
by squaring the first equation and replacing $b^2$. This can be written as 
$$c^3=(a^4b^5)k_3$$
where $k_3=b^2k_1^2k_2^3$ is an integer. Hence, $a^4b^5 \mid c^3$.
