Show that if a and b are positive integers, and $a^3 | b^2$, then a | b Show that if a and b are positive integers, and $a^3 | b^2$, then a | b.
If p is a prime divisor of a, and $p^r$ is the highest power of p dividing a. Then $p^{3r} | a^3$, and so $p^{3r} | b^2$. If $p^s$ is the highest power of p dividing b, then 3r ≤ 2s. And so we have, r ≤ (2/3)s < s.
But how come this implies that $p^r | b$, and eventually a | b.

Let $a,b$ be elements of $Z^+$  (i.e., the positive integers) and suppose $a^3|b^2$.
Then, by definition $b^2=c*a^3$ for some $c$ such that $c$ is an element of $Z^+$.
Hence, $\sqrt{b^2} = b = \sqrt{c*a^3} = (\sqrt{a^2})*(\sqrt{c*a}) = a*\sqrt{c*a}$
This works for both roots of $b^2$ and $a^2$.  
Consider case-by-case:
$\sqrt{b^2}$ = $b$ & $\sqrt{a^2}$ = $a$:   $b = a*(\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2}$ = $a$:  -$b$ = $a*(\sqrt{c*a}) \Rightarrow  b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ =$ b$ & $\sqrt{a^2} = -a:  b = -a*(\sqrt{c*a}) = a*(-\sqrt{c*a})$
$\sqrt{b^2}$ = -$b$ & $\sqrt{a^2} = -a: -b = -a*(\sqrt{c*a}) \Rightarrow b = a*(\sqrt{c*a})$
Hence, $a|b$ by definition.
