This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 530): Find the degree of $\alpha:=1+\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q}$
My efforts:
I first try to find the minimal polynomial by writing $\alpha=1+\sqrt[3]{2}+\sqrt[3]{4}\implies\alpha-1=\sqrt[3]{2}(1+\sqrt[3]{2})\implies(\alpha-1)^{3}=2(1+\sqrt[3]{2})^{3}$ but I didn't manage to get the minimal polynomial from this (which is, according to Wolfram, of degree $3$).
I also tried another method that failed: I noted that $\mathbb{Q}(\alpha)\subset\mathbb{Q}(\sqrt[3]{2})$ hence is of degree $\leq3$, moreover, since it is a subfield and $3$ is prime it only remains to show that $\alpha$ is not rational (which I can't prove).
Can someone pleases help me show that the degree is $3$ (preferably is one of the two methods I tried) ?