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

The 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 the unit cube's: the condition $x^3 = 2$ forces $x = \sqrt[3]{2}$, a root of $P(x) = x^3 - 2$. The problem was already ancient by the 5th century BCE: Hippocrates of Chios reduced it to inserting two mean proportionals between $1$ and $2$ — lengths $x, y$ with $1/x = x/y = y/2$, which algebraically says $x^2 = y$ and $y^2 = 2x$, hence $x^4 = 2x$ and $x^3 = 2$.

That reduction matters because it shows the Delian problem is really a question about a single real number's status, not about geometry at large: everything hinges on whether the specific length $\sqrt[3]{2}$ can be produced by the two classical tools. Wantzel's 1837 paper in Liouville's journal answers it by the same method he applies to angle trisection in the same memoir — translate constructibility into an algebraic degree condition, then check whether the required number satisfies it.

The remaining steps carry this out: Step 1 proves the general criterion that every constructible number has degree a power of $2$ over $\mathbb{Q}$; Step 2 shows $x^3 - 2$ is irreducible, so $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$; Step 3 collides the two facts. This is the more compact of Wantzel's two classical impossibilities, since only one number needs testing.