First proof (requires knowledge in fields extensions): the cubic rational extension is of degree 3 since is the minimal polynomial of (why?), and

a basis for it is precisely these three elements are linearly independent over , and this is precisely what had

to be proved.

Second proof: .

Note that (I assume ; if some of these two numbers is zero I leave to you to do the little changes necessary to fix the proof).

Raise to the 3rd power the expression , and now make some order here and deduce a contradiction to p not being a

perfect (rational, of course) cube.

Tonio