Note that $xy^2=a^3$ implies $x^2y=(xy/a)^3$.

Suppose $x=p_1^{e_1}\cdot p_2^{e_2}\cdot\cdot\cdot p_k^{e_k}$ and $y=p_1^{f_1}\cdot p_2^{f_2}\cdot\cdot\cdot p_k^{f_k}$ where $p_1,p_2,...,p_k$ are primes. If $xy^2$ is a perfect cube, then $e_i+2f_i\equiv 0 \pmod 3$. Multiplying this congruence by $2$, $2e_i+f_i\equiv 0 \pmod 3$ which is the desired. QED

Let $d=gcd(x,y)$. Then, $x=ad$ and $y=bd$ , where $gcd(a,b)=1$. We then have that $xy^2=d^3ab^2$ and $x^2y=d^3a^2b$. Now it suffices to prove the case where $d=1$. Since $gcd(a,b)=1$, we have that both $a$ and $b^2$ are cubes $\Rightarrow$ $a$ and $b$ are cubes. This obviously implies the desired result.

We have $log(x) +2log(y) =3log(a)$ for some $a\in N$ $3log(x) +3log(y) =3log(b)$ Now from these 2 equation we can get that $2log(x) +log(y) = 3log(c) $ for some $C \in N$

We see y² is a perfect square. therefore x= a³y, where a is any integer. Now xy² gives us the value of (ay)³. Now x²y = (a³y)².y = a⁶y².y =a⁶y³ =(a²y)³ Now a² and y are both Integers. Hence x²y is a perfect cube.

kapsitis wrote: Given positive integers $x$ and $y$ such that $xy^2$ is a perfect cube, prove that $x^2y$ is also a perfect cube. Is very easy with p-adic evaluation $v_p(xy^2)=v_p(k^3)$ where k is an integer so we have $v_p(x)$+$2v_p(y)$=$3v_p(k)$ and $2v_p(x)$+$v_p(y)$=$m$ where m is an integer We want to prove that m is a multiple of 3 But we will have 3$v_p(x)$+3$v_p(y)$=$3v_p(k)$+$m $ This implies that 3 devides m , what we wanted to prove