China Mathematical Competition

Denote the edge lengths of the three cubes as $a$, $b$ and $c$, respectively. Then we have $$6(a^2+b^2+c^2)=564,$$ i.e. $a^2+b^2+c^2=94$. We may assume that $$1\le a\le b\le c<10.$$

Then $$3c^2\ge a^2+b^2+c^2=94.$$ It follows that $c^2>31$. So $6\le c<10$, and this means that $c$ can only be $9$, $8$, $7$ or $6$.

If $c=9$, then $$a^2+b^2=94-9^2=13.$$ It is easy to see that $a=2$, $b=3$. So we get the solution $(a,b,c)=(2,3,9)$.

If $c=8$, then $$a^2+b^2=94-8^2=30.$$ This means that $b\ge 4$ and $2b^2\ge 30$; it follows that $b=4$ or $5$, so $a^2=5$ or $14$; in both cases $a$ has no integer solution.

If $c=7$, then $$a^2+b^2=94-7^2=45.$$ It is easy to see that $a=3$, $b=6$ is the only solution.

If $c=6$, then $$a^2+b^2=94-6^2=58.$$ So $2b^2\ge 58$, or $b^2\ge 29$. This means that $b\ge 6$, but $b\le c=6$, so $b=6$. Then $a^2=22$ and $a$ cannot be an integer.

In summary, there are two solutions: $(a,b,c)=(2,3,9)$ and $(a,b,c)=(3,6,7)$. Then the possible volumes are $$V_1=2^3+3^3+9^3=764{\text{cm}}^3,$$ $$V_2=3^3+6^3+7^3=586{\text{cm}}^3.$$