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.
 A: By hypothesis $\ n = \dfrac{a^3\!+b^3}{a^2b} = \dfrac{a}b + \left(\dfrac{b}a\right)^2\! =\, x+x^{-2}\,\overset{\large {\times\, x^2}}\Longrightarrow\,x^3-n\,x^2 + 1 = 0$
By the Rational Root Test $\ a/b\, =\, x\, = \pm 1\ \ $ QED
Generally applying RRT as above yields the degree $\,j+k\,$ homogeneous generalization
$$a,b,c_i\in\Bbb Z,\,\  a^{\large j}b^{\large k}\mid \color{#c00}1\:\! a^{\large j+k}\! + c_1 a^{\large j+k-1} b + \cdots + c_{\large j+k-1} a b^{\large j+k-1}\! + \color{#c00}1\:\!b^{\large j+k}\Rightarrow\, a = \pm b \qquad $$
$\qquad\qquad\ \ \ \ \ \  $ e.g. $\ a^2b \mid a^3 + c_1 a^2b + c_2 ab^2 + b^3\,\Rightarrow\, a = \pm b $ 
Alternatively the statement is homogeneous in $\,a,b\,$ so we can cancel $\,\gcd(a,b)^{\large j+k}$ to reduce to the case  $\,a,b\,$  coprime. The dividend $\,c\,$  has form $\,a^{\large n}\!+b^{\large n}\! + abm\,$ so by Euclid it is coprime to $a,b$ thus $\,a,b\mid c\,\Rightarrow\, a,b = \pm1$.
Remark $\ $ The proof in lhf's answer is precisely the standard proof of the Rational Root Test specialized to this particular polynomial. The Rational Root Test concisely encapsulates all divisibility results of this (homogeneous) form.
A: Hint: Here, $a,b$ can be negative.  It suffices to prove that $|a|=|b|$ when $\gcd(a,b)=1$ and $a^2b\mid a^3+b^3$.
A: 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 $B=1$.
A: 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 $a=b=\gcd(a,b)$.
