Show that 2^(1/3) + 3^(1/3) is irrational.by jebodh Tags: algebraic, construct, irrational, polynomial, roots 

#1
Mar2609, 12:26 PM

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 



#2
Mar2609, 02:01 PM

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 [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. 



#3
Mar2609, 02:55 PM

P: 2

Thanks, I'll try it now.
I can't believe I didn't do that. 


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