11.127. A cube with an edge length of $a$ is cut at the corners by planes such that a regular octagon remains from each face. Determine the volume of the resulting polyhedron.
## Solution.

Let the segment cut off by each of the specified planes on the edge of the cube be $x: A P=A M=E A_{1}=x$ (Fig. 11.23). Then $M P=E P=x \sqrt{2}$; $A P+P E+E A_{1}=A A_{1}, 2 x+x \sqrt{2}=a, \quad x=\frac{a}{2+\sqrt{2}}=\frac{a(2-\sqrt{2})}{2}$. We will calculate the volume of each of the cut-off pyramids $V_{1}=\frac{1}{6} x^{3}$.

Then the volume of the resulting polyhedron is the difference between the volume of the cube

$$
\text { and } \begin{aligned}
8 V_{1}: & V=a^{3}-8 V_{1}=a^{3}-\frac{4}{3} x^{3}=a^{3}-\frac{a^{3}(2-\sqrt{2})^{3}}{6}=\frac{a^{3}\left(6-(2-\sqrt{2})^{3}\right)}{6}= \\
& =\frac{a^{3}(6-8+12 \sqrt{2}-82+2 \sqrt{2})}{6}=\frac{a^{3}(14 \sqrt{2}-14)}{6}=\frac{7}{3} a^{3}(\sqrt{2}-1) .
\end{aligned}
$$

Answer: $\quad \frac{7}{3} a^{3}(\sqrt{2}-1)$.