Prove $ \sqrt{1 + \sqrt[3]{2}} $ is irrational using the theorem about rational roots of a polynomial 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. 
 A: 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, $(x^2-1)^3 = 2$.
A: 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.
A: 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 $x^3-2$.
A: 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.
A: 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.
