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
$$3c^2 \geq a^2 + b^2 + c^2 = 94.$$
It follows that $c^2 > 31$. So $6 \leq c < 10$, 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 $b \geq 4$ and $2b^2 \geq 30$; 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 $2b^2 \geq 58$, or $b^2 \geq 29$. This means that $b \geq 6$, but $b \leq c = 6$, 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
$$V_1 = 2^3 + 3^3 + 9^3 = 764\ \text{cm}^3,$$
$$V_2 = 3^3 + 6^3 + 7^3 = 586\ \text{cm}^3.$$