How can I show $\sqrt{2-\sqrt{2}} \in \mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$?

Question

How can I show $\sqrt{2-\sqrt{2}} \in \mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$?

I know $2+\sqrt{2} \in \mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$ and so I know $\sqrt{2} \in \mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$. Thus, I know $2-\sqrt{2}\in \mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$.

If $\alpha=\sqrt{2+\sqrt{2}}$, a basis for $\mathbb Q\Big(\sqrt{2+\sqrt{2}}\Big)$ over $\mathbb Q$ is $1, \alpha, \alpha^2, \alpha^3$. But I am not sure if/how I should use this.

Answer 1

Note that $\sqrt{2-\sqrt{2}}=\dfrac{\sqrt{2}}{\sqrt{2+\sqrt{2}}}$ and $\dfrac{1}{\sqrt{2+\sqrt{2}}}\in{\bf{Q}}(\sqrt{2+\sqrt{2}})$.