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> oh, geometrically? No. Algebraically works cleaner, and for any arbitrary positive solutions for r.

I hope you're not being serious. Just in case you are, your algebra is wrong. I'm quite certain you didn't look up what "doubling the cube" means, since the (faux) algebraic solution is y = cube_root(2 * x^3). It undercuts the rest of your comment.

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Doubling the cube was being ridiculous to prove the bad language to explain the problem.

If it really meant doubling the volume of a cube of X unit size, then absolutely it's (2 * x^3)^(1/3)

I don't think the author was trying to give a precise description of the problem. "Doubling the cube" is a term of art. It's like if he used the word "derivative" and you thought it meant a cheap copy of something, and then went further to prove how silly calculus was because of your misunderstanding of the term.

You're also selling the problem short. Doubling the cube is about producing a finite algorithm (given a limited set of operations) that realizes the value of (2 * x^3)^(1/3) concretely. An algebraic solution does not do this, because it stops at the inability to realize, say, the cube root of 2 explicitly.

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