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Please help me find the result of this example: $\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt 3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt 3}}}$

Thanky very much for your help

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  • $\begingroup$ Hint: just keep using the fact that $(a-b)(a+b)=a^2-b^2$. Group your terms accordingly. $\endgroup$ – lulu Oct 16 '16 at 10:37
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$\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt 3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt 3}}}$ $=\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt 3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt 3}}}=\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt 3}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt 3}}\right)}=\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot\sqrt{\left(2^2-\sqrt{2+\sqrt{2+\sqrt 3}}^2\right)}=$ $=\sqrt{2+\sqrt 3}\cdot\sqrt{2+\sqrt{2+\sqrt 3}}\cdot \sqrt{2-\sqrt{2+\sqrt 3}}=\sqrt{2+\sqrt 3}\cdot\sqrt{\left(2+\sqrt{2+\sqrt 3}\right)\cdot \left(2-\sqrt{2+\sqrt 3}\right)}=\sqrt{2+\sqrt 3}\cdot\sqrt{\left(2^2-\sqrt{2+\sqrt 3}^2\right)}=\sqrt{2+\sqrt 3}\cdot\sqrt{2-\sqrt 3}=\sqrt{(2+\sqrt 3)\cdot (2-\sqrt 3)}=\sqrt{2^2-\sqrt 3^2}=\sqrt 1=1$

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Denote $c=\sqrt{2+\sqrt{2+\sqrt{3}}}$ and multiply the last two radicals on the right to get

$$\sqrt{2+c}\cdot\sqrt{2-c}=\sqrt{4-c^2}=\sqrt{2-\sqrt{2+\sqrt{3}}}$$

Now repeat with $d=\sqrt{2+\sqrt{3}}$

$$\sqrt{2+d}\cdot\sqrt{2-d}=\sqrt{4-d^2}=\sqrt{2-\sqrt{3}}$$

And so the expression given is equal to

$$\sqrt{2+\sqrt{3}}\cdot\sqrt{2-\sqrt{3}}=1$$

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