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Show that 2^(1/3) + 3^(1/3) is irrational.

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that 2^(1/3) + 3^(1/3) is irrational. Hint: show that [x][/0] = 2^(1/3) + 3^(1/3) is algebraic by constructing an explicit polynomial f(x) with integer coefficients such that f([x][/0]) = 0. Then prove that f(x) has no rational roots.
    Note:[x][/0] means x subscript zero


    2. Relevant equations
    [x][/0] = 2^(1/3) + 3^(1/3)


    3. The attempt at a solution
    First I solved for x^3 and got 5+3{6^(1/3)[2^(1/3) + 3^(1/3)]}.
    Then I did x^9=[x^3]^3=> (after some work I got=> 125(x^3) -500 + [2^(1/3) + 3^(1/3)]{15[6^(2/3)] + 162}
    Do I cube the expression again, or I'm I missing something?

    My question is what am I supposed to do with this term,
    [2^(1/3) + 3^(1/3)]{15[6^(2/3)] + 162}?

    Thanks,
    Daniel
     
    Last edited: Mar 26, 2009
  2. jcsd
  3. Mar 26, 2009 #2
    The hint suggests to find a polynomial with integer coefficients. You have the right idea cubing, but there's a little more.

    You found [tex]x_0^3=5+\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})[/tex].

    But you can simplify that even more, by substituting [tex]x_0[/tex] in for [tex]\sqrt[3]{2}+\sqrt[3]{3}[/tex], to get [tex]x_0^3=5+\sqrt[3]{6}x_0[/tex] Then try to get rid of the final cube root, and you have an integer polynomial.
     
  4. Mar 26, 2009 #3
    Thanks, I'll try it now.
    I can't believe I didn't do that.
     
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