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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..?

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    First expand both sides and see what you can cancel. – 2017-02-19
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    @астонвіллаолофмэллбэрг I did, but I what should I do next? – 2017-02-19
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    Excellent. Now, if $b=0$, then any $a$ works as you can check. Otherwise, $a^2b + 2b^2 + 3ab -3a^2 + a = 0$ is true. Group powers of $b$ together, and apply the quadratic formula treating $a$ as a constant. If you are unclear about this, get back, I'm right here! – 2017-02-19
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    if $b \neq 0,$ evidently $( -1, -1),$ $( 8, -10),$ $( 9, -21),$ $( 9, -6).$ – 2017-02-19
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    @астонвіллаолофмэллбэрг Thanks. I think I can finish to the end now. – 2017-02-19
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    Brilliant! Always willing to help. – 2017-02-19
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    @астонвіллаолофмэллбэрг Thanks a lot! – 2017-02-19
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    @KayK.You are welcome! – 2017-02-19
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    @KayK. Your discriminant is missing a $-8a$ term under the first radical. – 2017-02-19
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    @dxiv Thanks. I was puzzled but now all cleared. – 2017-02-19
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    You should post your solution as an answer (instead of as an edit to your question). That way your question can be marked as "answered". – 2017-02-19
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    @MorganRodgers Okay, I will do so. Thanks. – 2017-02-19
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    @MorganRodgers But I just learned that I can't accept a self answer within 2 days so it will be still unanswered for 2 days. I'll ask астонвіллаолофмэллбэрг to post this as an answer. – 2017-02-19
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    That's okay, it will show up as having an answer that has been upvoted. – 2017-02-19
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    @астонвіллаолофмэллбэрг Would you please post it as an answer? You can copy and paste from my answer below. And then I'll delete mine. Thanks! – 2017-02-19
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    @MorganRodgers Okay, thanks! – 2017-02-19
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    @KayK. Thank you for requesting me to do so.However, I think my job was to get you to your answer, which I have done . If you are "ordering" me to do, I certainly shall, but I would like to politely decline your request, on the grounds that reputation does not matter to me, what matters is the questioner's (that is, yours) satisfaction. Your question has been excellent, looking at the upvotes, great credit to you for that! – 2017-02-19
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    @астонвіллаолофмэллбэрг I understood. I appreciate your help and the way you did. Best. – 2017-02-19

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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 $n^2$. $$a^2-8a=n^2$$ $$(a-4)^2:=m^2=n^2+16$$ $$(m,n)=(\pm4,0),(\pm5,\pm3)$$ $$(a,b)=(8,-10),(-1,-1),(9,-6),(9,-21)$$