Algebraic simplification including square roots $\sqrt{2}\sqrt{3 + 2\sqrt{2}} $ This may be stupid, but how do I see that $$\sqrt{2}\sqrt{3 + 2\sqrt{2}} - 1 = 1 + \sqrt{2}$$
having only the left-hand side? 

To denest the nested radical $\sqrt{3+2\sqrt{2}}$, we have a useful formula. Namely

Given a radical of the form $\sqrt{X\pm Y}$, we can rewrite it into$$\sqrt{X\pm Y}=\sqrt{\frac {X+\sqrt{X^2-Y^2}}2}\pm\sqrt{\frac {X-\sqrt{X^2-Y^2}}{2}}\tag1$$
  With $X>Y$ and $X,Y\in\mathbb{R}$.

In $\sqrt{3+2\sqrt{2}}$, we have $X=3,Y=\sqrt8$. Using $(1)$, we get$$\sqrt{3+2\sqrt{2}}=\left(\frac {3+1}2\right)^{1/2}+\left(\frac {3-1}2\right)^{1/2}=\sqrt2+1\tag2$$
From here, can you continue?
