TITLE: Prove $ \sqrt{1 + \sqrt[3]{2}} $ is irrational using the theorem about rational roots of a polynomial
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.