• Support PF! Buy your school textbooks, materials and every day products via PF Here!

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

  • Thread starter jebodh
  • Start date
1. Homework Statement
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. Homework 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:
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.
 
Thanks, I'll try it now.
I can't believe I didn't do that.
 

Related Threads for: Show that 2^(1/3) + 3^(1/3) is irrational.

Replies
2
Views
1K
  • Posted
Replies
10
Views
2K
Replies
5
Views
4K
  • Posted
Replies
3
Views
1K
Replies
3
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top