Question Number 137563 by liberty last updated on 04/Apr/21
$${let}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={k} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} } \\ $$$${in}\:{terms}\:{of}\:{k} \\ $$
Answered by Rasheed.Sindhi last updated on 04/Apr/21
$${let}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={k} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} } \\ $$$$−−−−−−− \\ $$$$\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={k} \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)}{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }={k} \\ $$$$\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }=\frac{{k}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }=\frac{{k}}{\mathrm{2}}+\frac{\mathrm{2}}{{k}} \\ $$$$\frac{\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{4}} −{y}^{\mathrm{4}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{4}} −{y}^{\mathrm{4}} \right)}=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{k}} \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{8}} +{y}^{\mathrm{8}} \right)}{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{k}} \\ $$$$\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{k}} \\ $$$$\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{k}}+\frac{\mathrm{4}{k}}{{k}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({k}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{4}{k}\right)^{\mathrm{2}} }{\mathrm{4}{k}\left({k}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{k}^{\mathrm{4}} +\mathrm{24}{k}^{\mathrm{2}} +\mathrm{16}}{\mathrm{4}{k}\left({k}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$ \\ $$
Answered by bramlexs22 last updated on 04/Apr/21
$${The}\:{equality}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:{k} \\ $$$${implies}\:\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }\:=\:{k} \\ $$$${hence}\:\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }\:=\:\frac{{k}}{\mathrm{2}}\:.{and}\:\left(\frac{{x}}{{y}}\right)^{\mathrm{4}} =\:\frac{{k}+\mathrm{2}}{{k}−\mathrm{2}}\: \\ $$$${Therefore} \\ $$$$\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }\:+\:\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }\:=\:\frac{\mathrm{2}\left({x}^{\mathrm{16}} +{y}^{\mathrm{16}} \right)}{{x}^{\mathrm{16}} −{y}^{\mathrm{16}} } \\ $$$$=\:\mathrm{2}\left(\frac{\left(\frac{{k}+\mathrm{2}}{{k}−\mathrm{2}}\right)^{\mathrm{4}} +\mathrm{1}}{\left(\frac{{k}−\mathrm{2}}{{k}+\mathrm{2}}\right)^{\mathrm{4}} −\mathrm{1}}\right)=\:\mathrm{2}\left(\frac{\left({k}+\mathrm{2}\right)^{\mathrm{4}} +\left({k}−\mathrm{2}\right)^{\mathrm{4}} }{\left({k}+\mathrm{2}\right)^{\mathrm{4}} −\left({k}−\mathrm{2}\right)^{\mathrm{4}} }\right) \\ $$