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I'm having trouble with this specific problem at the moment. The theorem states that if $n/m$ is a rational root of a polynomial with integer coefficients, the leading coefficient is divisible by m and the free coefficient is divisible by n.

Using this theorem, I'm supposed to prove that $ \sqrt{1 + \sqrt[3]{2}} $ is irrational. I don't have any idea where to start on this one.

Any help or hints are appreciated.

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I'm coming to this thread late, but if you're still interested in (other examples of) this method, here's a handout I wrote for a honors high school precalculus class back in Fall 1998: Proving Irrationality by the Rational Roots Theorem. – Dave L. Renfro Jan 11 '13 at 21:43
up vote 6 down vote accepted

You want to use the rational root theorem.

Hint: Let $x= \sqrt{1 + \sqrt[3]{2}}$, then, $x^2 = 1+ \sqrt[3]{2}$, so $(x^2-1) = \sqrt[3]{2}$. Hence, $(x^2-1)^3 = 2$.

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So basically, I assume that the number is rational and then I construct a polynomial that has it as it's root the way you did it, just with the 2 subtracted. Then I check all possible n/m pairs of the resulting polynomial and conclude that our number isn't among them? Is this at least close to right? – Luka Horvat Jan 11 '13 at 0:10
There's no assuming that the number is rational. I'm just forming an equation where $x$ is a root. Since it's hard to guess what equation might work, I showed how to make such an equation. The rest is correct, as in Clayton's solution. – Calvin Lin Jan 11 '13 at 0:32

Using Calvin Lin's hint above, we can expand the polynomial to $$(x^2-1)^3-2=x^6-3x^4-3x^2-3=0.$$ The Rational Root Theorem implies that the only possible rational roots are $\{\pm3,\pm1\}$. Checking these values shows that no roots are rational. By construction of the polynomial, we know in particular that $\sqrt{1+\sqrt[3]{2}}$ is irrational.

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