From a cube with a surface area of $384 \mathrm{~cm}^{2}$, we have cut out a cuboid with a square base as shown in the figure. The volume of the resulting octagonal prism is equal to three quarters of the volume of the original cube. Calculate the surface area of the prism.

![](https://cdn.mathpix.com/cropped/2024_04_17_de44a46afb7b0ea030d0g-2.jpg?height=499&width=494&top_left_y=493&top_left_x=815)
Let $a$ be the length of the edge of the cube. Then we have

$$
6 a^{2}=384
$$

From this, it follows that the length of the edge of the cube is $8 \text{ cm}$. Since the volume of the prism is equal to three quarters of the volume of the cube, the volume of the cut-out cuboid is equal to one quarter of the volume of the cube. Furthermore, if the height of the cuboid (and the prism) is equal to the length of the edge of the the original cube, the area of the square base of the cuboid is equal to one quarter of the area of a face of the cube, i.e.,

$$
\frac{1}{4} \cdot 8 \cdot 8=16 \text{ cm}^{2}
$$

The square base of the cuboid has an edge length of $\sqrt{16}=4 \text{ cm}$. For the surface area of the prism, we have:

Base area: $S_{p}=8^{2}-4^{2}=64-16=48 \text{ cm}^{2}$

Lateral surface area: Let $x$ be the distance of the cut-out hole from one of the side faces of the cube. Then we have:

$$
S_{p l}=[8+8+8+x+4+4+4+(8-4-x)] \cdot 8=40 \cdot 8=320 \text{ cm}^{2}
$$

Total surface area: $S=2 \cdot S_{p}+S_{p l}=2 \cdot 48+320=416 \text{ cm}^{2}$.