Compute the degree of finite field extension 
Compute the degree $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}]$.

By the Degree Theorem we have $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}(\sqrt{3 + \sqrt{3}})] \cdot [\mathbb{Q}(\sqrt{3 + \sqrt{3}}) : \mathbb{Q}]$.
Now, $\alpha = \sqrt{3 + \sqrt{3}} \iff \alpha^{2} = 3 + \sqrt{3} \iff \alpha^{2} - 3 = \sqrt{3} \iff \alpha^{4} - 6 \alpha^{2} + 6 = 0$. Which means that $\alpha = \sqrt{3 + \sqrt{3}}$ is a zero of $m(x) = x^{4} - 6 x^{2} + 6$. Now, $m$ is monic and Eisenstein with prime $p = 3$. Hence, $m$ is irreducible over $\mathbb{Q}$ and is thus the minimal polynomial. Hence we have $[\mathbb{Q}(\sqrt{3 + \sqrt{3}}) : \mathbb{Q}] = \text{deg}(m) = 4$.
To compute $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}(\sqrt{3 + \sqrt{3}})]$, we need to find the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{3 + \sqrt{3}})$.
This is where I am stuck.
Perhaps there might be a different solution if for instance $\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) = \mathbb{Q}(\bullet)$, but I can't seem to find such a field.
Also, can someone give me some general strategy on how to solve such problems? Perhaps some tips or what to watch out for given your experience.
 A: Here's a way to do it.
First of all, consider both $[\mathbb{Q}(\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ and $[\mathbb{Q}(\sqrt{3} ): \mathbb{Q}]$. The first is $4$ as you have stated and the second is $2$, so this gives you the hint that $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ is a multiple of $4$.
I want to prove that it is exactly $4$. Therefore I have to prove that $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})]=1$. That is equivalent to checking that I can express $\sqrt{3}$ as an element of $\mathbb{Q}(\sqrt{3+\sqrt{3}})$. All the elements of $\mathbb{Q}(\sqrt{3+\sqrt{3}})$ are like this:
$$\mathbb{Q}\left(\sqrt{3+\sqrt{3}}\right) = \left\{ a+b\sqrt{3+\sqrt{3}} + c\left(\sqrt{3+\sqrt{3}}\right)^2: a, b, c \in \mathbb{Q}\right\} = \left\{ a+b\sqrt{3+\sqrt{3}} + 3c + c\sqrt{3}: a, b, c \in \mathbb{Q}\right\}$$
Choosing $a = -3$, $b = 0$ and $c = 1$ you get that
$$\sqrt{3} \in \mathbb{Q}\left(\sqrt{3+\sqrt{3}}\right)$$
Therefore, as $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})]=1$ we have that
$$[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}] = [\mathbb{Q}(\sqrt{3+\sqrt{3}} ): \mathbb{Q}] \cdot [\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})] = 4 \cdot 1 = 4$$
