How to show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ So as title says I wanna show $\sqrt{3-\sqrt2} \in \mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$

So I know that the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ is $x^{4}-6x^{2}+7$ the zeroes of which are $\pm \sqrt{3-\sqrt2}, \pm \sqrt {3+\sqrt 2}$.

So far I have got as far as:

$\sqrt{3-\sqrt2}\sqrt {3+\sqrt 2} = \sqrt7$.

Any tips on how to find $\sqrt7$ in terms of the minimal polynomial of $\mathbb Q \left[\sqrt {3+\sqrt 2}\right ]$ ?
Since $2,7$ and $2\times 7$ are all nonsquares
in $\mathbb Q$, we see that $K={\mathbb Q}[\sqrt{2},\sqrt{7}]$ has
degree $4$ over $\mathbb Q$. If we put 
$\theta=\sqrt{7}+\sqrt{3+\sqrt{2}}$, we have

$$
\theta^2+4=14+\sqrt{2}+2\sqrt{7\times(3+\sqrt{2})}=
\sqrt{2}+2\sqrt{7}\theta \tag{1}
$$

So $\theta$ is a root of the polynomial 
$P=X^2-(2\sqrt{7})X+4-\sqrt{2}$. The discriminant of $P$ is
$d=3+\sqrt{2}$. If $d$ were a square in $K$, we would have
$x,y\in{\mathbb Q}[\sqrt{7}]$ such that $(x+y\sqrt{2})^2=3+\sqrt{2}$,
hence $x^2+2y^2=3,2xy=1$. This yields $y=\frac{1}{2x}$,
$x^2+\frac{2}{4x^2}=3$, so $2x^4-6x^2+1=0$ and hence
$x^2=\frac{3\pm\sqrt{7}}{2}$ which can be seen to be impossible in
${\mathbb Q}[\sqrt{7}]$.

We have therefore shown that $d$ is a nonsquare in $K$, so
$[K(\theta):K]=2$. Let $L={\mathbb Q}(\sqrt{7},\theta)$. 
By (1) above, $L$ contains $\sqrt{2}=\theta^2-2\sqrt{7}\theta+4$,
so $L$ contains $K(\theta)$. We deduce $L=K(\theta)$ as the other 
inclusion is trivial. Thus $[L:{\mathbb Q}]=8$.

Finally, if we had $\sqrt{3+\sqrt{2}}\in M$ where 
$M={\mathbb Q}[\sqrt{3-\sqrt{2}}]$, we would have
$\sqrt{7}\in M$ as explained in the OP, so $\theta\in M$
and eventually $M=L$, contradicting $[L:{\mathbb Q}]=8$.
Let us try this. Suppose $\alpha=\sqrt{3+\sqrt{2}}$. Then $\alpha^{2}=3+\sqrt{2}$, and $\alpha^{4}=11+6\sqrt{2}=11+6(\alpha^{2}-3)$ and hence $\alpha$ is a root of the polynomial $p(X)=X^{4}-6X^{2}+7$. This polynomial is irreducible, simply by attempting quadratic formula. Thus the degree $[\mathbb{Q}(\sqrt{3+\sqrt{2}}):\mathbb{Q}]=4$. Now suppose $\beta=\sqrt{3-\sqrt{2}}\in\mathbb{Q}(\sqrt{3+\sqrt{2}})$. Then

\begin{equation}
[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\beta)][\mathbb{Q}(\beta):\mathbb{Q}]
\end{equation}

Using the procedure similar for $\alpha$, you can show that $[\mathbb{Q}(\beta):\mathbb{Q}]=4$. Therefore $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]$, i.e. $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$. But any field isomorphism $\phi$ that maps $\alpha$ to $\beta$ cannot work: suppose $\phi(\alpha)=\beta$ up to $\pm$ sign.
\begin{equation}
\phi(\alpha)^{2}=3+\sqrt{2}=\beta^{2}=3-\sqrt{2}\Rightarrow \sqrt{2}=-\sqrt{2}
\end{equation}
a contradiction. Therefore our supposition is false.