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

by jebodh
Tags: algebraic, construct, irrational, polynomial, roots
 P: 2 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
 P: 114 The hint suggests to find a polynomial with integer coefficients. You have the right idea cubing, but there's a little more. You found $$x_0^3=5+\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})$$. But you can simplify that even more, by substituting $$x_0$$ in for $$\sqrt[3]{2}+\sqrt[3]{3}$$, to get $$x_0^3=5+\sqrt[3]{6}x_0$$ Then try to get rid of the final cube root, and you have an integer polynomial.
 P: 2 Thanks, I'll try it now. I can't believe I didn't do that.

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