Register to reply

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

Share this thread:
Mar26-09, 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}?

Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Mar26-09, 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.
Mar26-09, 02:55 PM
P: 2
Thanks, I'll try it now.
I can't believe I didn't do that.

Register to reply

Related Discussions
Proof: x is irrational => sqrt(x) is irrational Introductory Physics Homework 3
How to show show this proof using MAX Calculus & Beyond Homework 0
Proving an irrational to an irrational is rational Precalculus Mathematics Homework 3
I'll show you my google if you show me yours General Discussion 49
Irrational + irrational = rational Calculus 38