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 $\displaystyle \mathbb{Q}(\sqrt[3]{p})/\mathbb{Q}$ is of degree 3 since $\displaystyle x^3-p$ is the minimal polynomial of $\displaystyle \sqrt[3]{p}$ (why?), and

a basis for it is precisely $\displaystyle \{1,\,\sqrt[3]{p},\,\sqrt[3]{p^2}\}\Longrightarrow $ these three elements are linearly independent over $\displaystyle \mathbb{Q}$ , and this is precisely what had

to be proved.

Second proof: $\displaystyle a+b\sqrt[3]{p}+c\sqrt[3]{p^2}=0\iff (a+b\sqrt[3]{p})^3=$ $\displaystyle -c^3p^2\iff a^3+3a^2b\sqrt[3]{p}+3ab^2\sqrt[3]{p^2}+b^3p=-c^3p^2$ $\displaystyle \Longrightarrow \sqrt[3]{p}(a+b\sqrt[3]{p})=\frac{-c^3p^2-a^3-b^3p}{3ab}=q$ .

Note that $\displaystyle q\,,\,-c^3p^2\in\mathbb{Q}$ (I assume $\displaystyle ab\neq 0$ ; 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 $\displaystyle \sqrt[3]{p}(a+b\sqrt[3]{p})=q\Longrightarrow p(-c^3p^2)=q^3$ , 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 $\displaystyle a,b,c,p ~\neq 0 $

$\displaystyle a+ b \sqrt[3]{p} + c \sqrt[3]{p}^2 = 0 $ or

$\displaystyle a\sqrt[3]{p} + b\sqrt[3]{p}^2 + cp = 0 $

Let $\displaystyle x = \sqrt[3]{p} ~, ~ y= \sqrt[3]{p}^2 $

We obtain a system of linear equations :

$\displaystyle bx+cy = -a $

$\displaystyle ax + by = -cp $

Since $\displaystyle a,b,c,p$ are rational numbers , we must have

$\displaystyle {x,y} \in \mathbb{Q}$ . However, as you mentioned $\displaystyle p $ 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 ($\displaystyle \mathbb{Q}$ 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