Finding the degree of $1+\sqrt[3]{2}+\sqrt[3]{4}$ over $\mathbb{Q}$ 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) ?
 A: Let $\beta=\sqrt[3]2$. Then $\alpha=1+\beta+\beta^2$, hence $\mathbb Q(\alpha)\subseteq \mathbb Q(\beta)$. Clearly, the minimal polynomial of $\beta$ is $X^3-2$, hence $[\mathbb Q(\beta):\mathbb Q]=3$. As a vector space, $\mathbb Q(\beta)$ has $1, \beta, \beta^2$ as a basis, hence $\alpha$ is irrational (the only way to write $\alpha=a+b\beta+c\beta^2$ with rational coefficients is $\alpha=1+\beta+\beta^2$). Thus $1<[\mathbb Q(\alpha):\mathbb Q]|3$, i.e. and $[\mathbb Q(\alpha):\mathbb Q]=3$.

You can find the minimal polynomial of $\alpha$ itself:
$$\alpha = 1+\beta+\beta^2$$
$$\alpha^2 = (1+\beta+\beta^2)^2=1+2\beta+3\beta^2+2\beta^3+\beta^4 = 5+4\beta+3\beta^2$$
$$\alpha^3 = (1+\beta+\beta^2)(5+4\beta+3\beta^2)=5+9\beta+12\beta^2+7\beta^3+3\beta^4=19+15\beta+12\beta^2$$
Find a combination that eliminates all $\beta$ and $\beta^2$:
$$\alpha^2-3\alpha=2+\beta$$
$$\alpha^3-4\alpha^2=-1-\beta$$
$$\Rightarrow\quad \alpha^3-3\alpha^2-3\alpha-1=0$$
A: Denote $\alpha = 1 + \sqrt[3]{2} + \sqrt[3]{4}$. Since $\alpha \in \mathbb{Q}(\sqrt[3]{2})$, we get $\mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt[3]{2})$. Now 
$\sqrt[3]{2} + \sqrt[3]{4} \in \mathbb{Q}(\alpha)$
$(\sqrt[3]{4} + \sqrt[3]{2})^2 = 2 \sqrt[3]{2} + \sqrt[3]{4} + 4 \in \mathbb{Q}(\alpha)$
Subtracting the first element from the second implies $\sqrt[3]{2} \in \mathbb{Q}(\alpha)$, and therefore $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt[3]{2})$. 
A: Your second method is the right one. To see that $\alpha$ is not rational, note that
$$\frac{a}{b}=\alpha\implies \frac{a-b}{b\sqrt[3]{2}}=(1+\sqrt[3]{2})\implies \frac{(a-b)^3}{2b^3}=3\alpha\implies(a-b)^3=6ab^2$$
and note that this last equation is homogeneous, so if it has a solution it in $\mathbb Z$ it has a solution with $a$ and $b$ coprime. But $6|(a-b)^3$ so $6|(a-b)$, and so $6^2|ab^2$ so either $6|a$ or $6|b$, but either way since $6|(a-b)$ we get $6|a$ and $6|b$, hence $a$ and $b$ are not coprime. Thus no solution exists, so $\alpha$ is irrational.
A: If $1+ \sqrt[3]{2} + \sqrt[3]{4}=r$ with $r$ rational, then $\sqrt[3]{2}$ would satisfy $x^2+x+1-r=0$, contradicting the fact that is has degree $3.$
A: If $\alpha$ is rational, then so is $(2^{1/3} + 4^{1/3})$, and so the cube of that expression is rational; expanding, we see that the cube of that expression looks like a rational number plus $3*(32)^{1/3} = 6*4^{1/3}$; standard arguments tell you that this is irrational.
