Let $L=\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha)$ with $\alpha=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$
Show that:
i) $\alpha$ is a primitive element
ii) $L/\mathbb{Q}$ is a Galois extension.
iii) $L/\mathbb{Q}\simeq Q_8$
iv) There is no irreducible polynomial of the form $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$, which has a root in $L$
Next I put my solution omitting some calculation details and topics such as proving that the extensions obtained by each root are in effect automorphisms.
Let $K:=\mathbb{Q}(\sqrt{2},\sqrt{3})$
i)
Affirmation 0 . $\alpha$ is a primitive element of $L/\mathbb{Q}$
Proof: I did something similar to what is in this topic $\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$
ii)
Affirmation 1. $\alpha^2$ is not a square in $K$.
Proof. If $\alpha^2=c^2$ some $c\in K$. Let $\varphi\in \text{Aut}(K/\mathbb{Q})$ given by $\varphi(\sqrt{2})=\sqrt{2}$ and $\varphi(\sqrt{3})=-\sqrt{3}$ then $\alpha^2\varphi(\alpha^2)=6(2+\sqrt{2})^2$ therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$ Because $c\in K$ then $c=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ then $c\varphi(c)=(a+b\sqrt{2})^2-3(c+d\sqrt{2})^2\in\mathbb{Q}(\sqrt{2})$ Therefore $c\varphi(c)\in\mathbb{Q}(\sqrt{2})$ Therefore $6=\left(\frac{c\varphi{c}}{2+\sqrt{2}}\right)^2$ with $\frac{c\varphi{c}}{2+\sqrt{2}}\in \mathbb{Q}(\sqrt{2})$ Therefore $\sqrt{6}\in\mathbb{Q}(\sqrt{2})$ a contradiction.
Therefore $\alpha\not\in K$
Afirmation 2. $[L:\mathbb{Q}]=8$
Indeed, $[L:K][K:\mathbb{Q}]=[L:K][\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q}]$ and $[L:K]=2$ with $\text{Irr}_{K,\alpha}(x)=x^2-\alpha^2$, $[\mathbb{Q}(\sqrt{3})(\sqrt{2}):\mathbb{Q}(\sqrt{3})]=2$ with $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=x^2-2$ and $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$ with $\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=x^2-3$
Affirmation 3. $|\text{Aut}(L/\mathbb{Q})|=8$.
Indeed, by afirmation 2, $\text{Irr}_{K,\alpha}(x)=0$ implies $x\in \left\{\pm \alpha\right\}$ (in $L$) therefore exists two extension $\alpha\mapsto \pm \alpha$ Analogously, $\text{Irr}_{\mathbb{Q}(\sqrt{3}),\sqrt{2}}(x)=0$ implies $x\in\left\{\pm \sqrt{2}\right\}$ therefore exists two extension $\sqrt{2}\mapsto \pm \sqrt{2}$
$\text{Irr}_{\mathbb{Q},\sqrt{3}}(x)=0$ implies $x\in\left\{\pm \sqrt{3}\right\}$ therefore exists two extension $\sqrt{3}\mapsto \pm\sqrt{3}$
Therefore exists 8 $\mathbb{Q}$-automorphisms, this are: $\alpha\mapsto \pm \sqrt{ (2+\pm\sqrt{2})(3\pm\sqrt{3})}$
Therefore $|\text{Aut}(L/\mathbb{Q})|=8=[L:\mathbb{Q}]$ therefore $L/\mathbb{Q}$ is a Galois extension.
Why $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$, $ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) are in $L$?
This is because
$\beta=\frac{\alpha \beta}{\alpha}$ and $\alpha\beta=\sqrt{2}(3+\sqrt{3})\in \mathbb{Q}(\sqrt{2},\sqrt{3})\subset L$ then $\alpha\beta\in L$ therefore $\beta\in L$ etc.
iii)
Affirmation 4. $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.
Indeed, let $\varphi_{\beta},\varphi_{\gamma}\in \text{Gal}(L/\mathbb{Q})$ given by $\varphi_{\beta}(\alpha)=\beta$ and $\varphi_{\gamma}(\alpha)=\gamma$ with $\beta:=\sqrt{ (2-\sqrt{2})(3+\sqrt{3})},\gamma:=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ then $\varphi_{\beta}\circ \varphi_{\gamma}\neq \varphi_{\gamma}\circ \varphi_{\beta}$ therefore $\text{Gal}{L/\mathbb{Q}}$ is non abelian, this implies is not cyclic.
Moreover, $\varphi_{-\alpha}(\alpha):=-\alpha$ is the unique $\mathbb{Q}$-automorphism with order 2.
Because any group not cyclic and with a unique element of order $2$ is isomorphism to $Q_8$, then $\text{Gal}(L/\mathbb{Q})\simeq Q_8$.
To proves that iv) I am somewhat lost. I put below what I know about the minimal polynomial of $\alpha$.
Because $Irr_{L,\alpha}(x)=x-\alpha$ then
$Irr_{\mathbb{Q},\alpha}$ $=\prod_{\sigma\in \text{Gal}(L/\mathbb{Q})}(x-\sigma(\alpha))$ $=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)(x-\gamma)(x+\gamma)(x-\delta)(x+\delta)$
with $\pm\alpha=\pm\sqrt{ (2+\sqrt{2})(3+\sqrt{3})},\, \pm \beta=\pm\sqrt{ (2-\sqrt{2})(3+\sqrt{3})}$, $ \pm\gamma=\pm\sqrt{ (2+\sqrt{2})(3-\sqrt{3})},\, \pm\delta=\pm\sqrt{ (2-\sqrt{2})(3-\sqrt{3})}$ (conjugates of $\alpha$) (this is a knows results)
By contradiction, if $p(x)=x^4-ax^2+b\in\mathbb{Q}[x]$ has a root in $L$ then I should get a contradiction but I can't think of how.
How can I demonstrate this last part iv)? thank you
Actualization 1. I fix part iv)
iv) If $p(x):=x^4-b$ (irreducible add) with $b\in\mathbb{Q}$ is not a square show that $p(x)$ has not root in $L$.