10. In a sealed cubic box with an edge length of 6, there is a small ball with a radius of 1. If the box is flipped arbitrarily, the volume of the space that the ball cannot reach is $\qquad$ ;
When the box is arbitrarily flipped, the space that the small ball cannot reach is: the space at the 8 corners of the cube plus the space along the 12 edges of the cube. Among them, the space at the 8 corners can be combined into a cube with an edge length of 2, from which a sphere with a radius of 1 is removed, and its volume is $8-\frac{4}{3} \pi$; the space along the 12 edges can be combined into 3 hollow square prisms (a square prism with a base side length of 2 and a height of 4, from which a cylinder with a base radius of 1 and a height of 4 is removed), and its volume is $3(16-4 \pi)$. Therefore, the volume of the space that the small ball cannot reach is $8-\frac{4}{3} \pi+3(16-4 \pi)=56-\frac{40 \pi}{3}$.