<s>[INST] Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$ 
Find all pairs of nonzero integers $(a,b)$ such that
  $(a^2+b)(a+b^2)=(a-b)^3$

My effort
Rearranging the equation I have 
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we have the polynomial $Q(x)$ such that $$Q(x)=x^2(b^2+3b)+x(-3b^2+b)+2b^3 $$
So I must have that the roots $a_1,a_2$ must be such that 
\begin{array}
\space a_1+a_2=-\cfrac{-3b+1}{b+3} \\
a_1\cdot a_2=\cfrac{2b^2}{b+3} \\
\end{array}

I am stuck now,should I go by brute force and verify some values or
  there's still something I can do here to simplify the problem ?

 [/INST] The solutions of a quadratic with integer coefficients are either both rational or both irrational. So if one of $a_1,a_2$ is an integer the other must be rational. Say $a_1\in \Bbb Z$ and $a_2=\frac nm\in\Bbb Q$ is a rational with denominator $m$. Then the denominator of $a_1+a_2$ is also $m$ and the denominator of $a_1a_2$ is a divisor of $m$. We conclude that the denominator of $a_1a_2$ divides both $b+3$ and $b+1$, hence divides $2$.
Then $2a_1a_2$ is an integer, i.e., 
$$ b+3\mid 4b^2.$$


*

*If $3\nmid b$, $b+3$ and $b$ are coprime, hence $b+3\mid 4$ and $b\in\{-7,-5,-4,-2,-1,1\}$

*If $3\mid b$, say $b=3c$, we find $c+1\mid 12c^2$. As $c+1$ and $c$ are coprime, $c+1\mid 12$ and $b\in\{-39,-21,-15,-12,-9,-6,0,3,6,9,15,33\}$


The rest of the problem is finite.
</s>