## Task B-3.7.

A cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge $a$ is intersected by a plane passing through points $E \in \overline{A B}$, $F \in \overline{B C}$, and $G \in \overline{B_{1} C_{1}}$ such that $|A E|=\frac{1}{4}|A B|$, $|B F|=\frac{2}{3}|B C|$, and $\left|B_{1} G\right|=\frac{1}{3}\left|B_{1} C_{1}\right|$. The volume of the smaller of the two resulting geometric bodies is $\frac{7}{6}$. Determine the volume of the original cube.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-19.jpg?height=737&width=697&top_left_y=1833&top_left_x=608)

Notice that the resulting smaller geometric body is a truncated pyramid with height $a$.

Its lower base is a right triangle with legs $|E B|=\frac{3}{4} a$ and $|B F|=\frac{2}{3} a$, and the upper base is a right triangle with legs $\left|B_{1} G\right|=\frac{1}{3} a$ and $\left|H B_{1}\right|$.

Determine the length $x=\left|H B_{1}\right|$.

Consider the floor plan of the cube:

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-20.jpg?height=708&width=697&top_left_y=614&top_left_x=637)

From the similarity of the triangles, we have $x: \frac{3}{4} a=\frac{1}{3} a: \frac{2}{3} a$, from which we get $x=\frac{3}{8} a$.

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-20.jpg?height=948&width=374&top_left_y=1519&top_left_x=196)

Similarly, due to similarity, the height of the "entire" pyramid is $2a$.

The volume of the "entire" pyramid is

$V_{1}=\frac{1}{3}\left(\frac{1}{2} \cdot \frac{3}{4} a \cdot \frac{2}{3} a\right) \cdot 2 a=\frac{1}{6} a^{3}$.

The volume of the smaller pyramid (the complement) is

$V_{2}=\frac{1}{3}\left(\frac{1}{2} \cdot \frac{3}{8} a \cdot \frac{1}{3} a\right) \cdot a=\frac{1}{48} a^{3}$.

The volume of the truncated pyramid is

$V=V_{1}-V_{2}=\frac{1}{6} a^{3}-\frac{1}{48} a^{3}=\frac{7}{48} a^{3}$.

According to the given condition, $\frac{7}{48} a^{3}=\frac{7}{6}$, so $a^{3}=8$.

It follows that the volume of the cube is 8.