Question

\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)

Answer 1

\(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{8}}\cdot\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}=\sqrt{4+\sqrt{8}}\cdot\sqrt{4-2-\sqrt{2}}=\sqrt{4+\sqrt{8}}\cdot\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}=\sqrt{8+\sqrt{32}-\sqrt{32}-4}=\sqrt{4}=2\)

Answer 2

\(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+2\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2\left(4-2\right)}=\sqrt{2.2}=2\)