Firstly, what do we denote the polynomial ring over a field with an adjoint root? Do we denote it $\Bbb Q(\sqrt[3]{2})[x]$? I will for the following main question.
Why does $\Bbb Q(\sqrt[3]{2})=\{a+b\sqrt[3]{2} + c(\sqrt[3]{2})^2:a,b,c\in \Bbb Q\}$
Rather than just $a+b\sqrt[3]{2}$? When we are looking at $f(x)=x^3-2\in \Bbb Q(\sqrt[3]{2})[x]$ we have the root $f(\sqrt[3]{2})=0$, and I think this is where the problem sets in. Since this is a root, it should be reducible, and I get:
$$\frac{x^3-2}{x-\sqrt[3]{2}}= (x^2+\sqrt[3]{2}x+(\sqrt[3]{2})^2)$$ So we get: $x^3-2 = (x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+(\sqrt[3]{2})^2)$
So in factoring out our new root, we require that $(\sqrt[3]{2})^2$ is a root, else $x^3-2$ is still irreducible.
Is there a faster way to find what additional roots are required to adjoin a root 'properly'? I.e. I can't just say I want $q$ a root and take $Q(q)=\{a+qb|a,b\in \Bbb Z\}$ I might need some other things.
Typically I have seen it denoted $\mathbb{Q}[\sqrt[3]{2}]$.
If it is an extension over $\mathbb{Q}$, then we consider it a vector space with scalars being elements of $\mathbb{Q}$. Given that it needs to remain a field it must be closed under multiplication, and so $(a+b\sqrt[3]{2})(c+d\sqrt[3]{2})$ must be able to be generated by the basis. Since $(\sqrt[3]{2})^2\notin \mathbb{Q}$ we cannot get this element by means of multiplying by a scalar, so we must include it in our basis.
This depends on the element being adjoined