Impossibility of doubling the cube

Statement

The number $\sqrt[3]{2}$ is not a constructible number. So the cube cannot be doubled with straightedge and compass.

Remarks

The proof: $x^3 - 2$ is irreducible over $\mathbb{Q}$, by Eisenstein's criterion with $p = 2$. So $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3$. Since $3$ is not a power of $2$, the number $\sqrt[3]{2}$ is not constructible, by the degree theorem.

History

Pierre Wantzel proved the impossibility in 1837.

Explanations

L1 human

The edge of the doubled cube is the cube root of $2$. Its minimal polynomial has degree $3$, but constructible numbers have degree a power of $2$.