3. A plane cuts a cube such that three edges of the cube with a common vertex are cut at points that divide these edges in the ratios $2: 1, 3: 1$, and $4: 1$, counting from the common vertex. Determine the ratio of the volumes of the bodies into which this plane divides the cube.
Solution. Let $a$ be the edge, and $V=a^{3}$ be the volume of the cube. The cut body is the pyramid SMNP. The volume $V_{1}$ of this pyramid will be calculated if we take the right triangle $S M N$ as the base, and the segment $S P=\frac{a}{4}$ as the height. We have

$$
V_{1}=\frac{1}{3} \cdot \frac{a}{2} \cdot \frac{a}{3} \cdot \frac{a}{4}=\frac{a^{3}}{72}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_6f06ce47627480a4d953g-3.jpg?height=380&width=389&top_left_y=1322&top_left_x=1094)

The volume $V_{2}$ of the remaining part of the cube is $V_{2}=V-V_{1}=\frac{71 a^{3}}{72}$. Clearly, $V_{2}: V_{1}=71: 1$