Question Number 192045 by Safiullah_21 last updated on 06/May/23
$$\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{16}}]{\mathrm{2}}\right)}×\left(\frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}_{} }}\right) \\ $$$$ \\ $$$$\Rightarrow_{} \frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)}\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)} \\ $$$$ \\ $$