Question 1205472
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Each face of the resulting polyhedron is a regular octagon.  View the regular octagon as a square with the corners cut off at 45 degree angles.<br>
If s is the edge length of the resulting polyhedron, then the edge length of the cube (or side length of the square before the corners are cut off) is made up of one segment of length {{{s}}} and two segments of length {{{s/sqrt(2)}}}, for a total length of {{{s(sqrt(2)+1)}}}.<br>
Since the edge length of the original cube is 6, the edge length of the resulting polyhedron is<br>
{{{s=6/(sqrt(2)+1)=6(sqrt(2)-1)/((sqrt(2)+1)(sqrt(2)-1))=6(sqrt(2)-1)/(2-1)=6(sqrt(2)-1)}}}<br>
The volume of the resulting polyhedron is the volume of the original cube, minus the volume of the 8 pyramids cut off the corners.<br>
The volume of the original cube is 6^3=216.<br>
Each of the pyramids cut off the corners of the cube can be viewed as a pyramid with a base of an isosceles right triangle with side length {{{s/sqrt(2)}}} and a height of {{{s/sqrt(2)}}}.<br>
The area of the base of each of those pyramids is (one-half base times height)<br>
{{{(1/2)(s/sqrt(2))^2=s^2/4}}}<br>
The volume of each of the eight pyramids is (one-third base area times height)<br>
{{{(1/3)(s^2/4)(s/sqrt(2))=(s^3*sqrt(2))/24}}}<br>
The total volume of the eight pyramids is<br>
{{{8((s^3*sqrt(2))/24)=s^3*sqrt(2)/3=((6(sqrt(2)-1))^3)(sqrt(2)/3)=(216(5sqrt(2)-7))(sqrt(2)/3)=72(10-7sqrt(2))=720-504sqrt(2)}}}<br>
So the volume of the resulting polyhedron is<br>
{{{216-(720-504sqrt(2))=504sqrt(2)-504=504(sqrt(2)-1)}}}<br>