Question 1118068
Wikipedia (and other sources) would give you a formula for the volume of a cuboctahedron as a function of the edge length,
but formulas are for computers, who cannot think for themselves.
Humans are supposed to be able to think using their own brains.
 
The volume of the cuboctahedron is the volume of the original cube minus the volume of the parts cut off.
To cut the corners that way,
you will cut along lines connecting midpoints of adjacent edges.
Each of the six faces of the cube will be cut,
leaving a square face for the cuboctahedron,
and losing four corner triangles, like this:
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
green(rectangle(-1,-1,1,1)),
green(rectangle(-1,-1,-0.9,-0.9)),
green(rectangle(0.9,0.9,1,1)),
green(rectangle(-1,1,-0.9,0.9)),
green(rectangle(1,-1,0.9,-0.9)),
line(-1,0,0,-1),line(-1,0,0,1),
line(1,0,0,-1),line(1,0,0,1)
)}}} .
The black lines are the 1-cm edges of the cuboctahedron,
which has {{{6*4=24}}} such edges separating
6 square faces that are portions of the cube faces
from 8 triangular "freshly cut" faces.
The Pythagorean theorem tells us that
the length of the green sides of those "lost" isosceles right triangular pieces of each cube face is
{{{green(x)}}}{{{cm}}} such that {{{green(x)^2+green(x)^2=1}}} , so {{{green(x)=green(sqrt("1 / 2"))=green(sqrt(2)/2)}}} ,
and the length (in cm) of the edge of the cube is {{{green(2x)=green(sqrt(2))}}} .
 
The volume of the cube, before it was butchered) was
{{{(green(sqrt(2)))^3=green(2sqrt(2))}}} {{{cubic}}} {{{cm}}} .
 
As for the pieces cut off,
the 4 that were resting on the table are still there,
sitting on a right triangular face.
If you rearrange them, joining all right angles,
you get a nice square-base Egyptian-style pyramid,
with side length {{{1}}}{{{cm}}} , and height {{{green(sqrt(2)/2)}}}{{{cm}}} .
You can do the same with the 4 corners cut off from the top face of the cube.
Those two pyramids, if you join them by the square bases form an octahedron.
The volume of that octahedron (2 pyramids), in cubic cm, is
{{{2(1/3)(base_side^2)(height)=2(1/3)(1^2)(green(sqrt(2)/2))=sqrt(2)/3}}} .
 
So, you have an octahedrom made from the cut-off pieces,
and a cuboctahedron whose volume, in in cubic cm, is
{{{2sqrt(2)-sqrt(2)/3=2sqrt(2)-(1/3)sqrt(2)=(2-1/3)sqrt(2)=(5/3)sqrt(2)=highlight(5sqrt(2)/3)}}} .