Gorgonzola has explained

how to carve a dodecahedron out of a cube. In that
approach there is a

dodecahedron sitting

*inside* a

cube. I'm going to show a different construction
of the (regular) dodecahedron which starts from a

cube
and constructs a dodecahedron on the

*outside*.

Begin with a cube with edges of length
the the golden ratio, which I'll denote by *t*.
Recall that

*
t*^{2} - t - 1 =0 (*)

Why does the golden ratio come into this? Well
if you think about a regular

pentagon with edge length
1. Then the length of a diagonal is

*t*.
A diagonal of the pentagon is a straight line
joining any two non-adjacent vertices. Here's a picture
(with slightly odd scaling)

1
____
1 / \ 1
/______\
t
\ /
1 \ / 1
\/

The dodecahdron construction is quite simple. On each square face
of the cube we put up a tent. The tent has four sides
two of them are isosceles triangles and two of them are
trapezoids. Here's a picture (not to scale)
of the types of tent side indicating
the edge lengths.

/\ _1__
1/ \1 1/ \1
/____\ /______\
t t

Looking down on a face of the cube from above the tent
looks like this

______
|\ /|
| \____/ |
| / \ |
|/______\|

When we put up a tent on a face of the cube adjacent to
the this one we do it so that the ridge of the new tent
is perpendicular to the ridge of our exisiting tent.
Here's a picture (to a slightly weird scale)
of two adjacent faces of the cube
indicating this.

______
|\ /|
| \____/ | <--- cube face
| / \ |
|/______\|
| |
| \ / |
| \ / |
| \/ |
| | | <--- cube face
| | |
| /\ |
| / \ |
| / \ |
|________|

Notice that at the edge of the cube where the two tents
meet we have a pentagon whose edges have length 1.
It's clear now that we have indeed constructed
a regular dodecahedron with a cube inscribed,
except for one small detail.
The pentagon

____
/ \
/______\
\ /
\ /
\/

has a fold line (along the edge of the cube)
and we really should ensure that that the pentagon is
not actually folded. In other words we have to check
that the trapezoid and the isosceles triangle
are in the

*same* plane. Let

*a* be the
angle that the trapezoid makes with the cube face
and let

*b* be the angle that the isosceles
triangle makes with the cube face. We must show that

*a+b=pi*.

We need some more notation.
Let *x* be the height of the tent and let
*y* be the height of the isosceles triangle.
So we have the triangles:

/| /| /|\
y / | x / | x 1 / | \ 1
/ | / | / y| \
/b__| /a__| /___|___\
(t-1)/2 t/2 t/2 t/2

By the

Pythagorean theorem we have

*
y*^{2} + (t/2)^{2} = 1 and *y*^{2} = x^{2} + ((t-1)/2)^{2}

If we eliminate

*y* and use (*)
we can deduce from these equations that

*t=1/2*.
It's now easy to compute that tan

*a = 1/t*
and tan

*b = 1/(t-1)*. So using (*) again we see that
tan

*a*.tan

*b = 1*. It follows that

*a+b = pi*, as required.

Thus we have constructed a dodecahedron with an inscribed cube.
Clearly there is some choice here, there are five cubes
inscribed in the dodecahedron.