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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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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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  • $\begingroup$ 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? $\endgroup$ – Luka Horvat Jan 11 '13 at 0:10
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    $\begingroup$ 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. $\endgroup$ – Calvin Lin Jan 11 '13 at 0:32
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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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Here is a twist on the use of the Rational Root Theorem.

$\sqrt{1 + \sqrt[3]{2}}$ is a root of $(x^2-1)^3-2$. The Rational Root Theorem implies that rational roots of a monic polynomial with integer coefficients must be integers.

But $1 < \sqrt[3]{2} < 2$ implies $1 < \sqrt 2 < \sqrt{1 + \sqrt[3]{2}} < \sqrt 3 < 2$ and so $\sqrt{1 + \sqrt[3]{2}}$ cannot be an integer.

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Note that $x = \sqrt{1 + 2^{\frac{1}{3}}}$ satisfies $(x^2 - 1)^3 - 2 = 0$. One may use the Eisenstein criterion to conclude that this polynomial is irreducible over the rationals. But if $x$ were rational this polynomial would have a linear factor with rational coefficients, so $x$ must be irrational.

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If $x\in\mathbb{Q}$ then $x^2-1\in\mathbb{Q}$, so it suffices to prove $\sqrt[3]{2}\notin\mathbb{Q}$. You can do this with the rational root theorem by considering the polynomial $x^3-2$.

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  • $\begingroup$ BTW the polynomial $x^6-3x^4+3x^2-3$ meets Eisenstein's Criterion. $\endgroup$ – DanielWainfleet Jan 12 '17 at 1:14

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