Answer: $120,125$.

Solution. There are essentially two cases for painting the faces of the large cube:

- three painted faces form a "P" shape;
- three painted faces share a common vertex.
![](https://cdn.mathpix.com/cropped/2024_05_06_634e5517f3bc14ef6482g-6.jpg?height=308&width=666&top_left_y=92&top_left_x=396)

In the first case, if the painted $1 \times 1 \times 1$ cubes are "cut off," a parallelepiped $4 \times 5 \times 6$ remains. Then the number of small cubes without red faces is $4 \cdot 5 \cdot 6 = 120$.

In the second case, if the painted $1 \times 1 \times 1$ cubes are "cut off," a cube $5 \times 5 \times 5$ remains. Then the number of small cubes without red faces is $2 \cdot 5 \cdot 5 = 125$.
Problem 6.6. Zhenya painted three faces of a white cube $6 \times 6 \times 6$ red. Then he sawed it into 216 identical small cubes $1 \times 1 \times 1$. How many of the small cubes could have no red faces? List all possible options.