Worked solution: Dehn's invariant: the tetrahedron and cube are not scissors-congruent (1900)

For a regular tetrahedron the dihedral angle at every one of its six edges equals $\theta = \arccos(1/3)$, proved irrational as a multiple of $\pi$ in Step 5, so $[\theta] \ne 0$ in $\mathbb{R}/\pi\mathbb{Q}$; since all six edges have the same positive length $\ell$, $\operatorname{Dehn}(\text{tetrahedron}) = 6\ell \otimes [\theta]$, and this is nonzero in $\mathbb{R} \otimes_{\mathbb{Q}} (\mathbb{R}/\pi\mathbb{Q})$ because $\ell \ne 0$ pairs with a nonzero class $[\theta]$ in a $\mathbb{Q}$-vector space setting where such a tensor cannot collapse to zero unless one of the two factors does.

Since $\operatorname{Dehn}(\text{cube}) = 0 \ne \operatorname{Dehn}(\text{tetrahedron})$ (Step 4 vs. this step) and $\operatorname{Dehn}$ is a dissection invariant (Step 3), no cube can be scissors-congruent to a regular tetrahedron of the same volume — answering Hilbert's third problem negatively and showing that, unlike in the two-dimensional Wallace–Bolyai–Gerwien theorem for polygons, volume alone does not determine scissors-congruence class in three dimensions.