Worked solution: Wantzel's algebraic impossibility proof for doubling the cube (1837)

The classical 'doubling the cube' (or Delian) problem asks for a straightedge-and-compass construction, starting from a unit segment, of the edge $x$ of a cube whose volume is twice that of a unit cube: $x^3 = 2$, so $x=\sqrt[3]{2}$, a root of $P(x) = x^3 - 2$. The problem was already ancient by the time of Hippocrates of Chios (5th century BCE), who reduced it to finding two mean proportionals between $1$ and $2$, and it resisted straightedge-and-compass solution for over two thousand years.

Wantzel's 1837 paper resolves it by the same method used for angle trisection: translate constructibility into an algebraic degree condition, then check whether $x=\sqrt[3]{2}$ satisfies it. The remaining steps carry this out: first the general criterion that constructible numbers have degree a power of $2$ over $\mathbb{Q}$, then the fact that $P(x)=x^3-2$ is irreducible over $\mathbb{Q}$, giving $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ — not a power of $2$.

Unlike the trisection problem, here there is only one relevant number to test, $\sqrt[3]{2}$ itself, which makes this the more compact of Wantzel's two classical impossibility proofs.