Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

Here is an extremely easy way of showing that $\sqrt{2} \in \Bbb{Q}(\sqrt{2} + \sqrt[3]{2})$. Write $\alpha = \sqrt{2} + \sqrt[3]{2}$. Then $(\alpha-\sqrt{2})^3 = 2$, so that

$$\alpha^3 - 3\alpha^2(\sqrt{2}) + 6\alpha -2\sqrt{2} = 2.$$

It follows that $\alpha^3+ 6\alpha - 2 = \sqrt{2}(3\alpha^2+ 2)$, so that

$$\sqrt{2} = \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}.$$

Since $\Bbb{Q}(\alpha)$ is a field, the expression on the right hand side is in here, so that $\sqrt{2}$ is in here. No using horrible row reduction to calculate anything! It follows that $\sqrt[3]{2}$ is in here. Hence $\Bbb{Q}(\alpha)$ contains the fields $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}(\sqrt[3]{2})$ which means that $[\Bbb{Q}(\alpha) : \Bbb{Q}]$ is a multiple of 3 and 2. Since we already know that it is at most $6$, it follows now that since $2$ and $3$ are coprime that $[\Bbb{Q}(\alpha) : \Bbb{Q}] = 6$. Whatever monic polynomial of degree 6 that you find for $\alpha$ over $\Bbb{Q}$ will then be irreducible, and hence will be the minimal polynomial of $x$ over $\Bbb{Q}$.

There's a really easier way to do this. Notice that over $\mathbb Q[\sqrt 2]$, $\sqrt 2 + \sqrt[3]2$ is a root of $(x-\sqrt 2)^3 - 2$. The roots of this polynomial are $\sqrt 2 + \omega \sqrt[3]2$ with $\omega^3 = 1$, and clearly none of these are in $\mathbb Q[\sqrt 2]$ because if they would, we would have $\omega \sqrt[3]2 \in \mathbb Q[\sqrt 2]$ and the minimal polynomial of those three over $\mathbb Q$ is $x^3-2$, which is irreducible over $\mathbb Q$, hence $[\mathbb Q[\omega \sqrt[3]2] : \mathbb Q] = 3$, which means they can't be in $\mathbb Q[\sqrt 2]$ since $[\mathbb Q[\sqrt 2] : \mathbb Q] = 2$. Now knowing this, since a polynomial of degree $3$ is irreducible if and only if it has no roots, $(x-\sqrt 2)^3 - 2$ is also irreducible over $\mathbb Q[\sqrt 2]$, hence it is the minimal polynomial of $\sqrt 2 + \sqrt[3]2$ over $\mathbb Q[\sqrt 2]$ and $$ [\mathbb Q[\sqrt 2 + \sqrt[3]2] : \mathbb Q] = [\mathbb Q[\sqrt 2 + \sqrt[3]2] : \mathbb Q[\sqrt 2]][\mathbb Q[\sqrt 2] : \mathbb Q] = 3 \cdot 2 = 6. $$ Therefore you know that the minimal polynomial of $\sqrt 2 + \sqrt[3]2$ over $\mathbb Q$ has degree $6$. It now suffices to compute a polynomial of degree $6$ of which this guy is a root. That's computations : compute all powers of $\sqrt 2 + \sqrt[3]2$ up to the sixth power and use linear algebra. I see that you did, so I guess you're done.

And 1) is correct. You can use this fact to show that $2$ and $3$ divides the order of your degree extension, which means $6$ divides it. Since you found a poylnomial of degree $6$ of which your element is a root, the polynomial must be irreducible.

Hope that helps,

As noted in Patrick's answer, the degree of the minimal polynomial must be six. This given, it suffices to construct a polynomial of degree six with your number as a root. One way is brute force linear algebra. Here is a more efficient and systematic way; it is a part of a constructive proof that, given commutative rings $R \subseteq S$, the set of elements of $S$ which are integral over $R$ is a subring.

Let $\alpha$ and $\beta$ be algebraic numbers with conjugates $\alpha=\alpha_1,\alpha_2,\dots,\alpha_n$ and $\beta=\beta_1,\beta_2,\dots,\beta_m$. Then the polynomial

$$f(x)=\prod (x-\alpha_i-\beta_j)$$

has $\alpha+\beta$ as a root, and thanks to a symmetry argument, has coefficients in $\mathbb{Q}$. Evidently the degree of $f(x)$ is $mn$, which forces it to be the minimal polynomial if, as in your situation, you know for other reasons that $\alpha+\beta$ has degree $mn$. This gives a quick way to get your answer.

In regards to your question 4) above, it is a theorem that, given algebraic numbers $\alpha$ and $\beta$, "most" linear combinations $\alpha+\mu \beta$ (for $\mu \in \mathbb{Q}$) generate $\mathbb{Q}[\alpha,\beta]$. But to check whether this definitely happens for a particular choice may be time consuming (though you have to get "unlucky" for it to fail).