Is their an easier way to do this? 
I am trying to prove that:
$$\left[\mathbb Q\left(\sqrt 2, \sqrt 3, \sqrt{(9-5\sqrt 3)(2-\sqrt 2)}\right): \mathbb Q\right] = 8$$

Let $u = \sqrt{(9-5\sqrt 3)(2-\sqrt 2)}$.
We notice that $u^2 \in \mathbb Q(\sqrt 2, \sqrt 3) \implies \mathbb Q(u^2) \in \{\mathbb Q, \mathbb Q(\sqrt 2), \mathbb Q(\sqrt 3), \mathbb Q(\sqrt 6), \mathbb Q(\sqrt 2,\sqrt 3)\}$. The first four choices are clearly not possible, so $\mathbb Q(u^2) = \mathbb Q(\sqrt 2, \sqrt 3)$. Therefore, $\mathbb Q(\sqrt 2, \sqrt 3, u) = \mathbb Q(u)$.
A polynomial which has $u$ as a root is:
$$f(x) = x^8 -72 x^6 +720 x^4 -864 x^2 + 144$$
which can be found by repeated squaring and taking stuff to other sides in the equation $u = \sqrt{(9-5\sqrt 3)(2-\sqrt 2)}$. Hence $\left[\mathbb Q(u): \mathbb Q\right] \le 8$. Since $\mathbb Q(\sqrt 2,\sqrt 3)/\mathbb Q$ is a subextension of $\mathbb Q(u) /\mathbb Q$, $4$ divides $[\mathbb Q(u): \mathbb Q]$. So, $[\mathbb Q(u): \mathbb Q] = 4$ or $8$. If it is $4$, then so is the degree of $g(x)$, the minimal polynomial of $u$ over $\mathbb Q$. We know that $g(x) \mid f(x)$. This means that $f(x) = g(x) h(x)$ where $g,h \in \mathbb Q[x]$ are of degree $4$.

Lemma. Let $f(x) \in \mathbb Z[x]$ be monic and suppose $f(x) = g(x)h(x)$, $g(x),h(x) \in \mathbb Q[x]$. Then $f(x) = p(x)q(x)$ where $p(x), q(x) \in \mathbb Z[x]$ are primitive and $\deg p = \deg g, \deg q = \deg h$.

Proof
$\exists b_1,b_2 \in \mathbb Z$ such that $b_1 g(x), b_2 h(x) \in \mathbb Z[x]$. Let $\gamma = b_1 b_2$ and $A(x) = b_1 g(x)$, $B(x) = b_2 h(x)$. Then we have:
$$\gamma f(x) = A(x)B(x)$$
Let $a_1$ and $a_2$ be the contents of $A(x)$ and $B(x)$ respectively and $\delta = a_1 a_2$. Let $p_1(x) = A(x)/a_1$ and $q(x) = B(x)/a_2$. Then we have:
$$\gamma \delta^{-1} f(x) = p_1(x) q(x)$$
As $p_1(x),q(x)$ are primitive, it follows from Gauss' lemma that $\gamma \delta^{-1}f(x)$ is primitive, so $\gamma \delta^{-1} \sim 1$, i.e. $\gamma = \pm \delta$. Therefore: $f(x) = (\pm p_1(x))q(x)$. Let $p(x) = \pm p_1(x)$. Then $f(x) = p(x) q(x)$, $p(x),q(x) \in \mathbb Z[x]$ are primtive and degrees are as claimed.
$\square$
Hence $f(x) = p(x)q(x)$ where $p(x), q(x) \in \mathbb Z[x]$ and $\deg p = 4, \deg q = 4$.
If we write $p(x) = x^4 + ax^3 + bx^2 + cx + d$ and $q(x) = x^4 + a'x^3 + b'x^2 + c'x + d'$, then:
$$x^8 - 72 x^6 + 720 x^4 - 864 x^2 + 144 = x^8 + (a+a')x^7 + (b'+aa'+b)c^6 + (c' + ab'+ba'+c)x^5 + (ac'+bb'+ca'+d +d')x^4 + (ad' +bc' + cb' + da')x^3 + (bd'+cc'+db')x^2 + (cd'+dc')x + dd' $$
Then $c' = -c$, $a' = -a$ and $dd' = 144$. If $d = d' = 12$, we obtain a contradiction as follows:

*

*$c+c'+ab'+a'b = 0$, so $ab' - ab = 0$. Hence $a = 0$ or $b = b'$.


*If $a = 0$, then as $ad'+bc'+cb'+a'd =0$, $(b'-b)c = 0$, so $c = 0$ or $b=b'$.


*If $c=0$, then from $bd'+cc+db' = -864$ we get $(b+b')d = -864$, so $d \mid 864$ which is impossible. Hence $c \neq 0$, so $b=b'$.


*From $b' + aa' + b = -72$ we get $b = -36$. From $ac'+bb' + a'c + d = 720$ we get $b^2 + d=720$, so $d= 720 - 36^2$, which is impossible. Therefore, $a \neq 0$. so $b=b'$.


*We obtain the equations:
$$\begin{cases} 2b -a^2 = -72 \\ b^2 - 2ac = 708 \\ 24 b - c^2 = -864 \end{cases}$$
From the third equation we see that $c$ is divisible by $6$. From the second equation, $b$ is divisible by $6$ and from the first equation, $a$ is divisible by $6$. Writing $a = 6 \alpha$, $b=6 \beta$, $c = 6\gamma$, and substituting in the second equation, we get $36 \beta^2 - 72 \alpha \gamma = 708$. It follows that $708$ is divisible by $36$, which is not true.
"Similarly" we obtain a contradiction in the other cases ($d = d'=-12$, $d = 1$ and $d'=144$, etc.)
Therefore, $[\mathbb Q(u): \mathbb Q] = 8$.

Question. Is their an easier way to do this?

because this took a whole lot of work, and I didn't even work it out to the last details; I can't be truly sure that these cases lead to contradictions as well.
Before doing all of this, I thought about proving $u \notin \mathbb Q(\sqrt 2, \sqrt 3)$, but it turned out to be really difficult and I didn't know how to proceed.
 A: Here is one simple way to show that $(9 - 5 \sqrt{3}) ( 2- \sqrt{2})$ is not a square in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. Assume that we have an equality 
$$(9-5 \sqrt{3})(2-\sqrt{2}) = \alpha ^2$$
Consider the automorphism of that takes $\sqrt{2}$ to $- \sqrt{2}$ and fixes $\sqrt{3}$. We get 
$$(9-5 \sqrt{3})(2 + \sqrt{2}) = \alpha_1^2$$ and therefore
$$2=(2- \sqrt{2})(2+\sqrt{2}) = \left(\frac{\alpha \cdot\alpha_1}{9-5 \sqrt{3}}\right)^2$$ and so $\sqrt{2}$ lies in $\mathbb{Q}(\sqrt{3})$, not possible. 
A: All you have to do is to show that $v=(9-5\sqrt 3)(2-\sqrt2)$
is not a square in $K=\Bbb Q(\sqrt2,\sqrt3)$. If it were, write
$$u=\sqrt v=a+b\sqrt2+c\sqrt3+d\sqrt6.$$
Note that
$$\frac{2+\sqrt2}{2-\sqrt2}=\frac{(2+\sqrt2)^2}2=(1+\sqrt2)^2$$
and so
$$\sqrt{(9-5\sqrt3)(2+\sqrt2)}=(1+\sqrt2)(a+b\sqrt2+c\sqrt3+d\sqrt6)
=\pm(a-b\sqrt2+c\sqrt3-d\sqrt6)
=(a+2b)+(a+b)\sqrt2+(c+2d)\sqrt3+(c+d)\sqrt6.$$
So either $a+2b=a$ and $a+b=-b$ or $a+2b=-a$ and $a+b=-b$
(and corresponding stuff for $c$ and $d$). Thinking about
$\sqrt{(9+3\sqrt5)(2-\sqrt2)}$ gives equations linking $a$ and $c$
etc.
