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Question Number 192045 by Safiullah_21 last updated on 06/May/23

(1/((1+(√2))(1+(2)^(1/4) )(1+(2)^(1/8) )(1+(2)^(1/(16)) )))×(((1−(2)^(1/(16)) )/(1−(2_ )^(1/(16)) ))) ⇒_ ((1−(2)^(1/(16)) )/((1+(√2))(1+(2)^(1/4) )(1+(2)^(1/8) )(1−(2)^(1/8) )))⇒(1/((1+(√2))(1+(2)^(1/4) )(1−(2)^(1/4) )))

$$\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{16}}]{\mathrm{2}}\right)}×\left(\frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}_{} }}\right) \\ $$$$ \\ $$$$\Rightarrow_{} \frac{\mathrm{1}−\sqrt[{\mathrm{16}}]{\mathrm{2}}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{8}}]{\mathrm{2}}\right)}\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\left(\mathrm{1}+\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)\left(\mathrm{1}−\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)} \\ $$$$ \\ $$

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