<s>[INST] Suppose that $a$ and $b$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$. 
Suppose that $a$ and $b \in \mathbb{Z}^+$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.

I have reduced the above formulation to these two cases. Assuming $b = a + k$. Proving that any of the below two implies that $a=b$ will be enough.
$$a^2b|(a+b)^3 - 3ab^2$$
$$a^2b|2a^3+3a(a+k)+k^3$$
I can't proceed from here. How should I proceed from here?
Thanks.
 [/INST] Let $a=da_1$, $b=db_1$, where $d=\gcd(a,b)$, $a_1,b_1\in\mathbb Z^+$. Then $\gcd(a_1,b_1)=1$, $\gcd\left(a_1^2,b\right)=1$ and $$a^2b\mid a^3+b^3$$
$$\iff d^3a_1^2b_1\mid d^3a_1^3+d^3b_1^3$$
$$\iff a_1^2b_1\mid a_1^3+b_1^3$$
$$\iff \begin{cases}a_1^2\mid a_1^3+b_1^3\\b_1\mid a_1^3+b_1^3\end{cases}\iff \begin{cases}a_1^2\mid b_1^3\\ b_1\mid a_1^3\end{cases}$$
Since $\gcd\left(a_1^2,b_1^3\right)=\gcd\left(b_1,a_1^3\right)=1$ and $a_1,b_1\in\mathbb Z^+$, we get $a_1=b_1=1$. Therefore $a=b=\gcd(a,b)$.
</s>