Let $a,b\in\mathbb N$ such that $a\neq b$.
Consider $p,q\in\mathbb N$ such that $a=(a,b)p$ and $b=(a,b)q$. Then $(p,q)=1$ and $(p,(a,b))=1$. Hence $(p,q(a,b))=1$
If we define $c=p$ and $d=q(a,b)$ we have what we wanted.
CORRECTION:
Let $a,b\in\mathbb N$ such that $a\neq b$.
Consider $p,q\in\mathbb N$ such that $a=(a,b)p$ and $b=(a,b)q$. Then $(p,q)=1$ and $[a,b]=(a,b)pq$. Now we want to factorize $(a,b)$ in a suitable way, say $(a,b)=xy$, in order to get $c=xp$ and $d=yq$ coprime.
Let $s\in\mathbb{N}$ such that $(a,b)=(p,(a,b))\cdot(q,(a,b))\cdot s$. Notice that $(p,q)=1$ implies that $(p,(a,b))$ and $(q,(a,b))$ are coprime. Notice also that $s$ is coprime to p and q.
Now consider $c=p\cdot (p,(a,b))\cdot s$ and $d=q\cdot (q,(a,b))$. We have that $cd=[a,b]$, $c|a$ and $d|b$. The remaining part to show is that $c$ and $d$ are coprime.
Since $p$ and $q$ are coprime and $(q,(a,b))$ divides $q$, we conclude that $p$ is prime to $(q,(a,b))$ and then it is prime to $d$ as well.
Now, since $(p,(a,b))$ divides $p$, and $(p,d)=1$ then $(p,(a,b))$ is prime to $d$.
Similarly, since $s$ and $q$ are coprime, $s$ and $(q,(a,b))$ are coprime. Thus $s$ is prime to $d$.
Therefore $c=p\cdot (p,(a,b))\cdot s$ is prime to $d$.
Theorem (Bezut's Lemma): Let $a,b\in\mathbb{Z}$ such that $gcd(a,b)=d$. Then there exists $x,y\in\mathbb{Z}$ such that $ax+by=d$.
In your problem, we want all of these integers to be positive. Notice that $d\leq\min(a,b)$, so $d\leq |ax|,d\leq |by|$ and if $xy\neq 0$ at least one of $ax,by$ are negative. However, if both were the sum would be too, so exactly one is negative. WLOG let it be $by$ that is negative. We know that both $a$ and $b$ are positive by hypothesis, so $x$ is positive and $y$ is negative. Thus we can write $x=s$ and $y=-t$ for $s,t\in\mathbb{N}$. Thus $d=as-bt$.
Now for when $xy=0$. WLOG let $y=0$. In this case we get $d=ax$, so $d$ is a multiple of $a$. But $d$ is a divisor of $a$, so $d=a$. Thus $d=1a-0b$.
Best Answer
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.