$a^2 |b$ and $b^3 |c$ , prove $a^6 |c$

Question

Suppose $a$, $b$, $c$ ∈ Z. If $a^2 |b$ and $b^3 |c$ , prove $a^6 |c$

I have been asked this question, and am unsure how to solve. I have tried to equate $b = ma^2$ , but may have done this wrong.

Answer 1

Right, so let's continue from $b=ma^2$. Let's also set $c=kb^3$. Now substitute $b$... Do you see what happens there? Can you finish it?

Answer 2

If $a^2 | b$ then without loss of generality $\exists\, m \neq 0 : b=ma^2 $. Since $b^3 \ c$ then $\implies m^3a^6 | c$ or $a^6 |c$ since $m$ is arbitrary.