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 $[L:{\mathbb Q}]=8$.
Not elegant, but here is a direct approach to showing this is not true. Let $x=\sqrt{3+\sqrt{2}}$. Since $\{1,x,x^2,x^3\}$ is a basis for $\mathbb{Q}[x]$, then so is $$\{1,\sqrt{3+\sqrt{2}},\sqrt{2},\sqrt{6+2\sqrt{2}}\}$$ Suppose $$\sqrt{3-\sqrt{2}}=a+b\sqrt{3+\sqrt{2}}+c\sqrt{2}+d\sqrt{6+2\sqrt{2}}$$ with $a,b,c,d\in\mathbb{Q}$. Then squaring:
$$\begin{align} 3-\sqrt{2} &=a^2+b^2(3+\sqrt{2})+2c^2+d^2(6+2\sqrt{2})\\ &\phantom{{}={}}+2ab\sqrt{3+\sqrt{2}}+2ac\sqrt{2}+2ad\sqrt{6+2\sqrt{2}}\\ &\phantom{{}={}}+2bc\sqrt{6+2\sqrt{2}}+2bd(3+\sqrt{2})\sqrt{2}+4cd\sqrt{3+\sqrt{2}}\\ 3-\sqrt{2}&=(a^2+3b^2+2c^2+6d^2+4bd)\\ &\phantom{{}={}}+(b^2+2d^2+2ac+6bd)\sqrt{2}\\ &\phantom{{}={}}+(2ab+4cd)\sqrt{3+\sqrt{2}}\\ &\phantom{{}={}}+(2ad+2bc)\sqrt{6+2\sqrt{2}} \end{align}$$
From the latter two coefficients: $$ \left\{\begin{aligned} ab&=-2cd\\ ad&=-bc \end{aligned}\right. \implies \left\{\begin{aligned} abd&=-2cd^2\\ ad&=-bc \end{aligned}\right. \implies \left\{\begin{aligned} -b^2c&=-2cd^2\\ ad&=-bc \end{aligned}\right. $$ The top equation tells us either (1) $b^2=2d^2$ or (2) $c=0$. (1) is only possible if $b=d=0$, which would mean $\sqrt{3-\sqrt{2}}=a+c\sqrt{2}$, which would quickly lead to a contradiction. So if $c=0$:
$$\begin{align} 3-\sqrt{2}&=(a^2+3b^2+6d^2+4bd)\\ &\phantom{{}={}}+(b^2+2d^2+6bd)\sqrt{2}\\ &\phantom{{}={}}+(2ab)\sqrt{3+\sqrt{2}}\\ &\phantom{{}={}}+(2ad)\sqrt{6+2\sqrt{2}} \end{align}$$
Again looking at the last two coefficients, either (1) $b=d=0$ or (3) $a=0$. (1) has already been ruled out. So $a=0$, and:
$$\begin{align} 3-\sqrt{2}&=(3b^2+6d^2+4bd)\\ &\phantom{{}={}}+(b^2+2d^2+6bd)\sqrt{2} \end{align}$$
So $$ \left\{\begin{aligned} 3&=3b^2+6d^2+4bd\\ -1&=b^2+2d^2+6bd \end{aligned}\right. \implies \left\{\begin{aligned} 3&=3b^2+6d^2+4bd\\ -3&=3b^2+6d^2+18bd \end{aligned}\right. $$ Subtracting equations: $$\begin{align} 6&=-14bd\\ -\frac{3}{7}&=bd \end{align}$$ And lastly, using relations from above: $$ \begin{align} -1&=b^2+2d^2+6\left(-\frac{3}{7}\right)\\ \frac{11}{7}&=b^2+2d^2\\ 11&=7b^2+14d^2\\ 11b^2&=7b^4+14b^2d^2\\ 11b^2&=7b^4-6\\ 0&=7b^4-11b^2-6\\ 0&=(7b^2+3)(b^2-2)\\ \end{align}$$ And it's clear there is no such $b$ in $\mathbb{Q}$.
