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