Corners are sliced off a unit cube so that the six faces each become

If you're a rock star, come along for the ride! If not, take solace in the fact that I've never seen an official geometry question that's this tough and go find something else to do with your time!

Make one of the sides of the octagon s.

The legs of the 45-45-90s are \(\frac{s}{\sqrt{2}}\).

\(\frac{2s}{\sqrt{2}}+s = 1\)

\(\frac{2s}{\sqrt{2}}+\frac{s\sqrt{2}}{\sqrt{2}} = 1\)

\(2s+\sqrt{2}s=\sqrt{2}\)

\(s(2+\sqrt{2})=\sqrt{2}\)

\(s=\frac{\sqrt{2}}{2+\sqrt{2}}\)

\(s=\frac{1}{\sqrt{2}+1}\)

Multiply by \(\frac{\sqrt{2}-1}{\sqrt{2}-1}\)

\(s=\sqrt{2}-1\)

The volume of a pyramid is 1/3bh.

Let's make the base one of the sides that we can see. We already know h = \(s/\sqrt{2} = 1-\frac{1}{\sqrt{2}}\)

b = \(\frac{1}{2}(1-\frac{1}{\sqrt{2}})^2\)

b = \(\frac{1}{2}(1-\frac{1}{\sqrt{2}})(1-\frac{1}{\sqrt{2}})\)

b = \(\frac{1}{2}(1-\frac{2}{\sqrt{2}}+\frac{1}{2})\)

b = \(\frac{1}{2}(\frac{3}{2}-\frac{2}{\sqrt{2}})\)

b = \(\frac{1}{2}(\frac{3}{2}-\frac{2\sqrt{2}}{2})\)

b = \(\frac{3-2\sqrt{2}}{4}\)

1/3bh = \(\frac{1}{3}(\frac{3-2\sqrt{2}}{4})(1-\frac{1}{\sqrt{2}})\)

1/3bh = \(\frac{1}{3}(\frac{3-2\sqrt{2}}{4})(\frac{\sqrt{2}}{\sqrt{2}}-\frac{1}{\sqrt{2}})\)

1/3bh = \(\frac{1}{3}(\frac{3-2\sqrt{2}}{4})(\frac{2}{2}-\frac{\sqrt{2}}{2})\)

1/3bh = \(\frac{1}{3}(\frac{3-2\sqrt{2}}{4})(\frac{2-\sqrt{2}}{2})\)

1/3bh = \(\frac{6-3\sqrt{2}-4\sqrt{2}+4}{24}\)

1/3bh = \(\frac{10-7\sqrt{2}}{24}\)

There are 8 such pyramids, so \(\frac{10-7\sqrt{2}}{3}\)

Answer choice B.

Typing that all out in forum-friendly formatting was good times! But not great times...