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.

Write $d=\gcd(a,b)$ and $a=dA$, $b=dB$, with $\gcd(A,B)=1$.
Then we can cancel $d^3$ on both sides of $a^2b \mid a^3+b^3$ and get $A^2B \mid  A^3+B^3$.
This implies that $A^2 \mid B^3$ and $B \mid A^3$, and so $A=B=1$.
Indeed, $A^2 \mid B^3$ implies that every prime that divides $A$ divides $B$. Since $\gcd(A,B)=1$, we must have $A=1$. But then $B \mid A^3$ implies $B=1$.
