Find all positive integers x, y Find all positive integers x, y such that $(x^2 + y)(x + y^2) = (xy)^3$

I get the problem from a practice problem set of a math camp which was organised for our national math olympiad.

I think that I can solve it if I multiply the left but after multipling I found no other way other than give up.
 A: Your RHS is big, really big, and you might want to use that.
Assume WLOG that $x\geqslant y$. Then we have two cases.
Case 1: $x\geqslant y^2$. Then $(x^2 + y)(x + y^2)\leqslant(x^2 + x)\cdot2x\leqslant4x^3$, so $y^3\leqslant4$, so for it to be a cube, $y^3=1$, and $(x^2+1)(x+1)=x^3$, which can't happen.
Case 2: $y\leqslant x<y^2$. Then $(x^2 + y)(x + y^2)<(x^2 + x)\cdot2y^2\leqslant4x^2y^2$, so $xy<4$, which leaves a pretty short list of options for both, of which none works.
All in all, no solutions.
So it goes.
A: If $x,y$ are both odd, then $(x^2+y)(x+y^2)$ is even and $(xy)^3$ is odd.
If one of $x,y$ is even and the other is odd, then $(x^2+y)(x+y^2)$ is odd and $(xy)^3$ is even.
So $x,y$ must both be even and therefore $\geq 2$.  Hence $\dfrac{x^2}{y} \leq \dfrac{x^2}{2}$ and $1 < \dfrac{x^2}{2}$ so that:
$$\dfrac{x^2}{y} + 1 < \dfrac{x^2}{2}+\dfrac{x^2}{2} = x^2$$
$$x^2 + y < x^2y$$
Similar reasoning shows that $x+y^2 < xy^2$.  Hence:
$$(x^2+y)(x+y^2) < (x^2y)(xy^2) = (xy)^3$$
So there are no solutions in positive integers.
