1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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}?

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook