<s>[INST] 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? 
 [/INST] Try to write $2(3+2\sqrt 2)=6+4\sqrt 2$ as a square of $a+b\sqrt 2$ for some rational numbers $a,b$:
$$(a+b\sqrt 2)^2=a^2+2b^2+2ab\sqrt2 \;=\; 6+4\sqrt 2$$
By $\Bbb Q$-linear independence of $1$ and $\sqrt 2$, this implies $$ab=2,a^2+2b^2=6$$
Can you take it from here?

Here are some hints, if needed:

If we try to find integer solutions $a,b$, we easily find $a=2,b=1$ as solution.
 Therefore:
$$\sqrt 2 \cdot \sqrt{3+2\sqrt 2}= \sqrt{6+4\sqrt 2} = \sqrt{(2+\sqrt 2)^2}=2+\sqrt 2$$

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