Prove that set is a field. I should prove that the following is a field: $$\{a + b\sqrt[3]2 + c\sqrt[3]4: a, b, c \in \Bbb Q \}$$ 
Is it enough to prove inverses for + and * here?   I think I can manage to prove that $$\{a + b\sqrt7, a, b \in \Bbb Q \}$$ is a field (by showing inverses, identity and closure) but this one seems a lot more complicated. For instance what is the point of the last term? Why is it there? 

Thank you.
 A: The last term is there because it has to be. For example, if you didn't have the "$c\sqrt[3]{4}$", then how would you show that $(\sqrt[3]{2})^2$ is an element of your set?
And to answer your question, yes, you need to prove that addition and multiplication of any two elements in your set remain in your set. You need to show that they're associative, commutative, that there exists an additive identity and multiplicative identity. You need to then show that everything has an additive inverse, and everything but zero has a multiplicative inverse.
In other words, you need to verify that your set, together with the standard addition and multiplication satisfy the defining properties of a field, which can be found on wikipedia.
The associativity and commutativity come from the fact that your set is a subset of $\mathbb{C}$, and your addition and multiplication are restrictions of the operations on $\mathbb{C}$, where they are of course associative and commutative. Thus, your additive (resp. multiplicative) identity are 0 (resp. 1), which are easily seen to be in your set.
Thus, the only significant parts of the exercise are showing that your set is closed under addition and multiplication, and demonstrating inverses.
A: Since $\mathbb{F} = \{a + b\sqrt[3]2 + c\sqrt[3]4, a, b, c \in \Bbb Q \}$ is a subset of $\mathbb{C}$, it suffices, as you said, to prove that inverses for $+$ and $*$ are contained in $\mathbb{F}$.
The third term isn't there by accident - $\mathbb{F}$ is the smallest field containing both $\mathbb{Q}$ and $\sqrt[3]2$ (notice that $\sqrt[3]4 = (\sqrt[3]2)^2$); it is also a vector space over the field of rationals, $\mathbb{Q}$.
The fact that you need three terms to describe it is a restatement of the fact that $\mathbb{F}$ (as a vector space) has dimension $3$ over $\mathbb{Q}$.
Now, an interesting question is "how do we know that the dimension is three?". This is a consequence of $\mathbb{F}$ being generated by $\sqrt[3]2$. Notice the difference:  
We need two terms for $\sqrt 7$, and $\sqrt 7$ is a root of $x^2-7$ (a polynomial of degree $2$);
we need three terms for $\sqrt[3]2$, and $\sqrt[3]2$ is a root of $x^3-2$ (a polynomial of degree $3$).
These are just special cases of a general statement:

The dimension of $\mathbb{Q}(\alpha)$ - that is, the smallest field which contains both $\mathbb{Q}$ and $\alpha$ - is the degree of the (unique) irreducible monic polynomial $f\in\mathbb{Q}[X]$ such that $f(\alpha) = 0$.

A: Hint
Set $K=\bigl\{a+b\sqrt[3]2+c\sqrt[3]4\mid a,b,c\in\mathbf Q\bigr\}$. You have to checkit is stable by addition and multiplication, contains $0$ and $1$ hence it is a subring of $\mathbf C$.
For the existence of the inverse of an element $x\ne 0$, and consider multiplication by $x$ as an endomorphism of the $\mathbf Q$-vector space $K$. What pproperties does it have?
