Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $ 
Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $; equivalently, prove that $\mathbb Q (\sqrt 2, \sqrt 3, u)$ is of degree $2$ over $\mathbb Q (\sqrt 2, \sqrt 3) $ where $u $ is a number such that $u^2 = (9-5\sqrt 3)(2-\sqrt 2) $.

This question is a small section of what is required for this question: Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{(9 - 5\sqrt{3})(2-\sqrt{2})})$ is normal over $\mathbb{Q}$, and finding its Galois group.
The answer of the above question reads as follows:
 "Note that $Gal(\mathbb Q(\sqrt 2, \sqrt 3)/ \mathbb Q(\sqrt 3))$ has order $2$, generated by $σ_{1,0} ∈ Gal(\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q)$, where $σ_{1,0}$ fixes $\sqrt 3$ and sends $\sqrt 2$ to its negative. Let $N$ denote the norm of the extension $\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q(\sqrt 2)$; if $α^2 = x^2$ for some $x ∈ \mathbb Q(\sqrt 2, \sqrt 3)$, then $N(α^2) = N(x^2) = N(x)^2 ∈ \mathbb Q(\sqrt 3)$, i.e. $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$. We note $N(α^2) = (9 − 5√3)^2(2)$so if $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$, then so is $2$, a contradiction.
However, as a person totally unfamiliar with field norms, I was wondering if there is any other way to solve this, or at least explaining the similar idea of the proof in simpler notions. Brute-forcingly writing $u^2 = (a+b\sqrt2+c\sqrt3+d\sqrt6)^2$ with all letters as rationals and expanding it seems a lot tedius - may take hours to solve everything, and there's no gaurantee that this would lead to a contradiction. Standard result in Galois theory&Group theory are welcome.
 A: Here is a more "hands-on" (and systematic) approach. Suppose that 
$$u^2=(2-\sqrt{2})(9-\sqrt{3})\tag{1}$$
for some $u\in{\mathbb Q}(\sqrt{2},\sqrt{3})$. You can write $u=v+w\sqrt{3}$
with $v,w\in{\mathbb Q}(\sqrt{2})$. Expanding in (1) yields 
$$
(v^2+3w^2)+(2vw)\sqrt{3}=9(2-\sqrt{2})-(2-\sqrt{2})\sqrt{3} \tag{2}
$$
Since $1$ and $\sqrt{3}$ are linearly independent over ${\mathbb Q}(\sqrt{2})$, it follows that 
$$
v^2+3w^2=9(2-\sqrt{2}), \ 2vw = -(2-\sqrt{2}) \tag{3}
$$
The second equality yields $w=-\frac{2-\sqrt{2}}{2v}$, and reinjecting this into
the first equality we obtain 
$$
v^2+3\bigg(\frac{2-\sqrt{2}}{2v}\bigg)^2=9(2-\sqrt{2}) \tag{4}
$$
which can be rewritten as
$$
v^4-9(2-\sqrt{2})v^2+3\bigg(\frac{2-\sqrt{2}}{2}\bigg)^2=0 \tag{5}
$$
or
$$
v^4-9(2-\sqrt{2})v^2+3(1-\sqrt{2})=0 \tag{6}
$$
Complating the square, we find that this is equivalent to
$$
\bigg(v^2-\frac{9(2-\sqrt{2})}{2}\bigg)^2=78(\sqrt{2}-1) \tag{7}
$$
Writing $v^2-\frac{9(2-\sqrt{2})}{2}$ as $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$, as in the deduction
of (3) we obtain
$$
a^2+2b^2=-78, 2ab=78 \tag{8}
$$
and the first equality is impossible for rational $a,b$.
