59th Ukrainian National Mathematical Olympiad

Answer: $a=2,b=10$.

It can be seen from the problem statement that both unknowns are even, because otherwise we would have an equality of an even and an odd number. Let us denote $a=2n$ and $b=2k\Rightarrow$ $$2n\cdot 8k^3+8n^3+2k=2018\text{ or }8n{k}^3+4n^3+k=1009.$$ Here, it can be seen that $k$ must be odd, and now a simple exhaustive search will be enough to find the solution $n{k}^3\le \frac{1009}{8}$, hence, $k^3\le 125$. Therefore, there are only three options.

$k=5\Rightarrow 1000n+4n^3=1004\Rightarrow n=1$ satisfies the inequality. For larger values of $n$, the left-hand side becomes larger than $1004$.

$k=3\Rightarrow 64n+4n^3=1006\Rightarrow$ no solution exists, since 4 divides the left-hand side but not the right-hand side.

$k=1\Rightarrow 8n+4n^3=1008\Rightarrow 2n+n^3=252\Rightarrow n=2m\Rightarrow 4m+8m^3=252\Rightarrow m+2m^3=63$. Now, we need to check only three cases for $m$, since, from the last equation, $m\le 3$. $m=1,m=2$ and $m=3$ do not satisfy the equation.

Hence, there is only a single pair $k=5$ and $n=1\Rightarrow b=10$ and $a=2$.