If a,b,c and p are rational numbers,

and a+bp^1/3+cp^2/3=0

and p is not a perfect cube.

prove that a=b=c=0

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- May 3rd 2010, 08:02 AMjashansinghalfind it
If a,b,c and p are rational numbers,

and a+bp^1/3+cp^2/3=0

and p is not a perfect cube.

prove that a=b=c=0 - May 3rd 2010, 09:04 AMtonio

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 - May 4th 2010, 06:54 AMjashansinghalQuote:

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).

- May 5th 2010, 02:18 AMsimplependulum
Assume

or

Let

We obtain a system of linear equations :

Since are rational numbers , we must have

. However, as you mentioned is not a perfect cube , a contradiction . - May 6th 2010, 07:47 AMFancyMouse
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. - May 6th 2010, 08:35 AMgrgrsanjayconfused first but clear now
thanks for explaining