The Problems of Ukrainian Authors

Let's denote the edge of the big cube as $a$ and the edges of two not unit as $b$ and $c$. Then we have an equation: $$b^3+c^3+2008=a^3.$$ Because $12^3=1728<2008$, then $a>12$, so $a\ge 13$. Now let's find an upper bound of $a$. It is clear that $a\ge b+c$, $$b^3+3b^{2}c+3b{c}^2+c^3\le a^3=b^3+c^3+2008,$$ which means that $b^{2}c+b{c}^2\le \frac{2008}{3}$ or $b^{2}c+b{c}^2\le 669$. Then $2c^3\le b^{2}c+b{c}^2\le 669$ so $c^3\le 334$.

Moreover, $$\begin{gathered} 2b^2+8\le b^{2}c+b{c}^2\le 669\Rightarrow 2b^2\le 661\Rightarrow b^2\le 330\Rightarrow b\le 18\Rightarrow b^3\le 5832\Rightarrow \\ {b}^3+c^3\le 6166\Rightarrow a^3=b^3+c^3+2008\le 6166+2008=8174\Rightarrow a\le 20. \end{gathered}$$ The only thing that remains is to test on cube of integer numbers $b\ge c\ge 2$, which satisfy an equation $b^3+c^3+2008=a^3$, where $a$ is from $13$ to $20$. As a result, we have the unique solution: $5^3+4^3+2008=13^3$.