Is $ (3+\sqrt{2})^{2/3} $ an irrational number?

Question

I am supposed to find out whether $ (3+\sqrt{2})^{2/3} $ is an irrational number and prove it, but I have no idea how to begin. Thanks

Answer 1

Say it is rational, name it $q$. Then $$q^3 = (3+\sqrt{2})^2 = 11+6\sqrt{2}\implies \sqrt{2}= \underbrace{q^3-11 \over 6}_{\in\mathbb{Q}}$$

A contradiction.

Answer 2

Notice that $\alpha=(3+\sqrt{2})^{2/3}$ is a root of $X^6-22X^3+49$ and this polynomial has no rational roots, since such a root would be an integer that divides $49$ and none of them are a root.