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 ?


$b=0$ is a solution for any $a$.  We can then divide it out, looking for non-zero solutions.  If we just feed it to the quadratic equation, we get $$a=\frac {3b-1 \pm \sqrt{(1-3b)^2-8(b+3)b^2}}{2(b+3)}$$ You must have $b \lt 0$ to have the discriminant positive.  A quick search finds $(-1,-1), (9,-6), (8,-10), (9, -21)$ with no others by $b=-1000$
