doubling the cube
Legend says the people of Delos, begging the oracle to end a plague, were told to double the size of their cubic altar. They naively doubled every edge — but that makes the volume eight times larger, not two. The real task is to build a cube with exactly twice the volume of a given one, using compass and straightedge. Like angle trisection, it sounds plausible and is in fact impossible.
Doubling the volume means the new edge must be the old edge times the cube root of 2: if the original edge is 1, the new edge has length 2^(1/3), since (2^(1/3))^3 = 2. So the whole problem collapses to a single question: is the cube root of 2 a constructible number? Its minimal polynomial over the rationals is x^3 - 2, which is irreducible (it has no rational root). Therefore the cube root of 2 has degree 3 over the rationals. But every constructible number has degree a power of 2 — and 3 is not. So 2^(1/3) cannot be constructed, and the cube cannot be doubled with these tools.
Notice the family resemblance: trisecting a 60-degree angle, and doubling the cube, both founder on a degree-3 obstruction — a cubic that square-root constructions can never crack. Squaring the circle fails even harder, on the transcendence of pi. As always, the impossibility is tied to the unmarked straightedge and compass; with a marked ruler or by intersecting conics (a method known to the Greeks), the cube root of 2 can indeed be constructed.
A unit cube has volume 1; a cube of volume 2 needs edge 2^(1/3). But 2^(1/3) is a root of the irreducible cubic x^3 - 2, so its degree over the rationals is 3 — not a power of 2 — and it is not constructible.
Doubling volume needs the cube root of 2, a degree-3 number beyond the reach of square-root constructions.
Doubling the EDGE (which beginners try) gives eight times the volume, not two — that is the whole trap. The impossibility is for compass and straightedge only; intersecting conics or a marked ruler can produce the cube root of 2.