Here are some computations of Dehn invariants (which you might want to read before this node, if nothing makes much sense) in three dimensions. While I cannot recommend the computation of Dehn invariants as a particularly interesting activity, I should point out that the particular Dehn invariants computed here are of great relevance to the proof for Hilbert's third problem. Also that if you've come all this way, the end of the proof is really really close -- don't give up!

A cube
A cube C has all dihedral angles π/2. Let f:MCQ be any linear functional suitable for computing Dehn invariants. Then f(π/2)=0, so Df(C)=0.
R is a tetrahedron with each face having one dihedral right angle with another face (hence "quadri-rect-angular tetrahedron; you've gotta love the names given in Geometry during the 19th century...). This is any tetrahedron similar to the tetrahedron with vertices (0,0,0), (1,0,0), (1,1,0) and (1,1,1). We can assemble a cube from 6 orthoschemes. Since congruent polyhedra have equal Dehn invariants, and since Dehn invariants are "additive", we immediately have for any Dehn invariant Df that 6⋅Df(R)=Df(C)=0, so Df(R)=0. This is a tetrahedron with all Dehn invariants zero.
A tetrahedron with a nonzero invariant
By now, it may well seem that all Dehn invariants are zero, and our troubles have been for nothing. So here's a tetrahedron with a nonzero invariant. T is the right tetrahedron with vertices O(0,0,0), A(1,0,0), B(0,1,0) and C(0,0,1). It has 3 edges (OA, OB and OC) of length 1 and a dihedral angle of π/2. And it has 3 more edges (AB, BC and AC) of length sqrt(2) and some dihedral angle θ.

To discover more about θ, we note that it is precisely the angle between AD and OD, where D(0,1/2,1/2) is the midpoint between B and C. A bit of analytic geometry (just take the inner product between the two vectors A-D and O-D, and divide by their lengths) shows that cos θ=1/sqrt(3). And we immediately have that sin θ=sqrt(2/3).

Now we need a bit of number theory. First,

cos 2θ = cos2θ - sin2θ = 1/3 - 2/3 = -1/3
is a rational number. But cos is an almost frothy function -- see trigonometric number for the amazing details. The only rational multiple of π with a rational cosine is π/3 (with cosine 1/2). So 2θ, which has some other rational cosine, cannot be a rational multiple of π.

In other words, the set {&pi,θ} is linearly independent over the field of rational numbers. So there is some linear functional

f:{aπ + bθ: a,b∈Q}→R
which satisfies f(π)=0 and f(θ)=1.

The Dehn invariant of T with respect to f is Df(T) = 3⋅1⋅0 + 3⋅sqrt(2)⋅1 = 3⋅sqrt(2) ≠ 0.