If a,b,c and p are rational numbers,
and p is not a perfect cube.
prove that a=b=c=0
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.
When p is not a perfect square, one can use Eisenstein's criterion to prove x^3-p is irreducible. But it's nontrivial (at least I think) to show x^3-p is irreducible in general case, i.e. merely if p is not a perfect cube of some element in the ground field ( here).
And a warning if anyone want to generalize it: it's not true that if q is not a perfect 4-th power, then x^4-q is irreducible. e.g. if q=-4, then one can factor x^4+4=(x^2-2x-2)(x^2+2x-2)
EDIT: OK I forgot that a cubic polynomial is irreducible iff it doesn't have a root, so in this case it's trivial to show x^3-p is irreducible.