Estonian Mathematical Olympiad

If $b=0$, then according to the equation $2a^2+b=0$ from which $a=0$.

Assume now that $b\ne 0$. As $b^3$ and $(2a^2+b{)}^3$ are perfect cubes, their ratio $a$ is the cube of a rational number; as it is an integer, it is the cube of an integer $c$. By taking cubic root from each side of the equation we get a

relation $2c^6+b=bc$, implying $b(c-1)=2c^6$. As $c$ and $c-1$ are coprime, $c^6$ and $c-1$ are also coprime. Therefore, $c-1$ has to divide $2$ and $c$ must be one of $3$, $2$, $0$ or $-1$.

If $c=3$, then $a=27$ and we get $2b=2\cdot 729$, whence $b=729$.

If $c=2$, then $a=8$ and we get $b=128$.

If $c=0$, then we get $-b=0$, which has already been analysed.

If $c=-1$, then $a=-1$ and $-2b=2$ from which $b=-1$.