Question Number 207226 by hardmath last updated on 10/May/24
$$\mathrm{Find}:\:\:\:\frac{\left(\mathrm{2}\:−\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \:\centerdot\:\left(\mathrm{6}\:+\:\mathrm{4}\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }{\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} }\:=\:? \\ $$
Answered by Sutrisno last updated on 10/May/24
$$\left(\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}+\sqrt{\mathrm{2}}}.\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\left(\mathrm{4}−\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{2}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{4}−\mathrm{2}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}\right)\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{36}−\mathrm{32}}{\mathrm{2}}\right)^{\mathrm{8}} \\ $$$$\left(\frac{\mathrm{4}}{\mathrm{2}}\right)^{\mathrm{8}} =\mathrm{256} \\ $$
Answered by Rasheed.Sindhi last updated on 10/May/24
$$\:\left(\frac{\left(\mathrm{2}\:−\:\sqrt{\mathrm{2}}\right)\:\centerdot\:\left(\mathrm{6}\:+\:\mathrm{4}\:\sqrt{\mathrm{2}}\right)}{\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right)}\right)^{\mathrm{8}} \: \\ $$$$\:=\left(\frac{\left(\mathrm{2}\:−\:\sqrt{\mathrm{2}}\right)\:\centerdot\:\left(\:\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \right.}{\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right)}\right)^{\mathrm{8}} \: \\ $$$$=\left(\:\left(\mathrm{2}−\sqrt{\mathrm{2}}\:\right)\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right)\:\right)^{\mathrm{8}} \\ $$$$=\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{8}} =\mathrm{256} \\ $$
Commented by hardmath last updated on 10/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professors} \\ $$