After writing this solution, I wanted to see if I could locate a post which might corroborate my attempt. The closest (in approach) I could find was Kell Berliner's solution (2nd soln from the top down) posted here. I wish to receive feedback on whether my solution correctly addresses XRBtoTheMOON's question regarding how we know $\sqrt{ca}\in\mathbb{Z}$ (written in my solution as $\sqrt{ak}$). I believe this can be accomplished with the following parity argument.
Proof Attempt:
[SETUP]
Let $a,b\in\mathbb{Z}^+$ and suppose $a^3\mid b^2$. Hence $\exists k\in\mathbb{Z}$ st. $a^3k=b^2$.
And thus, $a^2(ak)=b^2$. Or equivalently, $a\sqrt{ak}=b$.
We claim $\sqrt{ak}\in\mathbb{Z}$.
Simple analysis reveals that $a^3k=b^2\implies a(a^2k)=b^2\implies a\mid b^2$.
The FTOA (fundamental theorem of arithmetic) tells us that the right-hand side & the left-hand side of an equality must have the same number of each type of prime factor. In particular, the $\color{blue}{\text{LHS}}$ and $\color{green}{\text{RHS}}$ of $\color{blue}{a^2(ak)}=\color{green}{b^2}$ must have the same number of each prime factor. As the $\color{green}{\text{RHS}}$ is a perfect square, $a^2(ak)$ will necessarily have an even number of each prime factor in its factorization. There are two types of primes to consider in the factorization of $a^2(ak)$: those primes $p$ st. $p\mid a$ and those primes $p$ st. $p\nmid a$.
[PARITY]
$\underline{\text{Case 1:}}$ Consider a prime $p$ st. $p\mid a$. Such a $p$ will either appear an even or odd number of times in the factorization of $a$. If $p$ appears an even number of times in $a$, then it must also appear an even number of times in the factorization of $k$ (if it all). If $p$ only appeared an odd number of times in $k$, then there'd be an odd number of factors of $p$ on the $\color{blue}{\text{LHS}}$ & an even number of factors of $p$ on the $\color{green}{\text{RHS}}$, contradicting the FTOA. Since $p$ appears an even number of times in both $a$ and $k$, the product $ak$ will contain an even number of factors of $p$. If $p$ appears an odd number of times in $a$, then it must appear an odd number of times in $k$. Again, if $p$ appeared an odd number of times in $a$ but only an even number of times in $k$, then the $\color{blue}{\text{LHS}}$ would have an odd number of factors of $p$ and the $\color{green}{\text{RHS}}$ would have an even number of factors of $p$ contradicting the FTOA. No matter the scenario we're faced with, $ak$ will necessarily contain an even number of a given prime $p$ which divides $a$.
$\underline{\text{Case 2:}}$ Consider a prime $p$ st. $p\nmid a$, but $p\mid b^2$. As $p$ must be present in even quantity on the $\color{blue}{\text{LHS}}$, but is in no way contributed to the factorization of the $\color{blue}{\text{LHS}}$ through $a$, it must be that $k$ contains all those factors of $p$ (if such primes exist). Ergo, the product $ak$ contains an even number of factors of $p$.
Between cases 1 & 2, I believe this establishes that, regardless of the status of a given prime (i.e. whether $p\mid a$ or $p\nmid a$), it's true $p$ will appear an even number of times in the factorization of $ak$. This means $ak$ is a perfect square and hence $\sqrt{ak}\in\mathbb{Z}^+$.
As $\sqrt{ak}\in\mathbb{Z}^+$, this means $a\mid b$ as desired.
$\blacksquare$
