Prove $ \sqrt{1 + \sqrt[3]{2}} $ is irrational using the theorem about rational roots of a polynomial

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.

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, $(x^2-1)^3 = 2$.

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.