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  1. #1
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    find 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
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  2. #2
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    Quote Originally Posted by jashansinghal View Post
    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


    First proof (requires knowledge in fields extensions): the cubic rational extension \mathbb{Q}(\sqrt[3]{p})/\mathbb{Q} is of degree 3 since x^3-p is the minimal polynomial of \sqrt[3]{p} (why?), and

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

    to be proved.

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

    Note that q\,,\,-c^3p^2\in\mathbb{Q} (I assume 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 \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
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  3. #3
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    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).
    would you please help in fixing it?
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  4. #4
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    Assume  a,b,c,p ~\neq 0

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


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

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

    We obtain a system of linear equations :

     bx+cy = -a
     ax + by = -cp

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

     {x,y} \in \mathbb{Q} . However, as you mentioned  p is not a perfect cube , a contradiction .
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  5. #5
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    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 ( \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.
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  6. #6
    Member grgrsanjay's Avatar
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    Cool confused first but clear now

    thanks for explaining
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