I have an abstract algebra proof I am a little iffy on. Help please! I was independently studying abstract algebra and found the following problem:
The problem says to show $\mathbb{Q}(\sqrt{3},\sqrt{7}) = \mathbb{Q}(\sqrt{3} + \sqrt{7})$.
The solution given was as follows:

Everything made since except line 4. Where does the $-3(\sqrt{3} + \sqrt{3})$ come from and why does it equal $4\sqrt{3}$?
Would someone be able to show be how this works for any case? In other words,
$\mathbb{Q}(\sqrt{a}, \sqrt{b}) = \mathbb{Q}(\sqrt{a} + \sqrt{b})$ where $gcd(a,b) = 1$. 
I wanted to use a similar argument as the above proof.
 A: $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b}) \Rightarrow \sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$.
Then, $\sqrt{ab}(\sqrt{a}+\sqrt{b}) = a\sqrt{b} + b\sqrt{a} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$ 
$\Rightarrow a\sqrt{b} + b\sqrt{a} - a(\sqrt{b}+\sqrt{a}) = b\sqrt{a}-a\sqrt{a} \in \mathbb{Q}(\sqrt{3}+\sqrt{7})$
$\Rightarrow \displaystyle\frac{1}{b-a} \cdot \sqrt{a}(b-a) = \sqrt{a} \in \mathbb{Q}(\sqrt{a} + \sqrt{b})$
Since $\sqrt{a} \in \mathbb{Q}(\sqrt{a} + \sqrt{b}), \sqrt{b} = (\sqrt{a} + \sqrt{b}) - \sqrt{a} \in \mathbb{Q}(\sqrt{a},\sqrt{b})$.
A: Hint: There are some trivial cases, when one of $a$ or $b$ is square. Otherwise, $\sqrt{ab}\notin\mathbb Q$ (why?). Then try mimicking the proof verbatim.
A: $\mathbb Q(\sqrt3+\sqrt7)$is a sub field of $\mathbb Q(\sqrt3,\sqrt7)$. The degree of the former must divide that of the later, which is clearly 4. The degree of $\mathbb Q(\sqrt3+\sqrt7)$ is greater than two, so must be 4, so the two fields are the same.
