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.