<s>[INST] Find $a, b\in \mathbb Z$ where $(a^2+b)(a+b^2)=(a-b)^3$ Find all non-zero $a, b\in \mathbb Z$ where
$$(a^2+b)(a+b^2)=(a-b)^3$$
I actually had no clue on what to try. Thanks for your help.

I believe I've already tried but per the 1st comment let me expand both sides and see what I cancel.
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3$$
$$b(a^2b+2b^2+3a^2-3ab+a)=0$$
And then..?
 [/INST] Let me post it as an answer to mark this answered.
Thanks again to астон вілла олоф мэллбэрг!
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3$$
$$b(a^2b+2b^2+3a^2-3ab+a)=0$$
$$b=0\quad or \quad 2b^2+(a^2-3a)b+3a^2+a=0$$
Applying quadratic formula on b,
$$b=\frac{a(3-a)\pm\sqrt{a^2(a-3)^2-24a^2-8a}}{4}=\frac{a(3-a)\pm (a+1)\sqrt{a(a-8)}}{4}$$
So $a(a-8)$ should be a square, let's say $n^2$.
$$a^2-8a=n^2$$
$$(a-4)^2:=m^2=n^2+16$$
$$(m,n)=(\pm4,0),(\pm5,\pm3)$$
$$(a,b)=(8,-10),(-1,-1),(9,-6),(9,-21)$$
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