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Question Number 137563 by liberty last updated on 04/Apr/21

let ((x^2 +y^2 )/(x^2 −y^2 ))+((x^2 −y^2 )/(x^2 +y^2 ))=k find the value of ((x^8 +y^8 )/(x^8 −y^8 ))+((x^8 −y^8 )/(x^8 +y^8 )) in terms of k

$${let}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={k} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} } \\ $$$${in}\:{terms}\:{of}\:{k} \\ $$

Answered by Rasheed.Sindhi last updated on 04/Apr/21

let ((x^2 +y^2 )/(x^2 −y^2 ))+((x^2 −y^2 )/(x^2 +y^2 ))=k find the value of ((x^8 +y^8 )/(x^8 −y^8 ))+((x^8 −y^8 )/(x^8 +y^8 )) −−−−−−− (((x^2 +y^2 )^2 +(x^2 −y^2 )^2 )/((x^2 +y^2 )(x^2 −y^2 )))=k ((2(x^4 +y^4 ))/(x^4 −y^4 ))=k ((x^4 +y^4 )/(x^4 −y^4 ))=(k/2) ((x^4 +y^4 )/(x^4 −y^4 ))+((x^4 −y^4 )/(x^4 +y^4 ))=(k/2)+(2/k) (((x^4 +y^4 )^2 +(x^4 −y^4 )^2 )/((x^4 +y^4 )(x^4 −y^4 )))=((k^2 +4)/(2k)) ((2(x^8 +y^8 ))/(x^8 −y^8 ))=((k^2 +4)/(2k)) ((x^8 +y^8 )/(x^8 −y^8 ))=((k^2 +4)/(4k)) ((x^8 +y^8 )/(x^8 −y^8 ))+((x^8 −y^8 )/(x^8 +y^8 ))=((k^2 +4)/(4k))+((4k)/(k^2 +4)) =(((k^2 +4)^2 +(4k)^2 )/(4k(k^2 +4))) =((k^4 +24k^2 +16)/(4k(k^2 +4)))

$${let}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }={k} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} } \\ $$$$−−−−−−− \\ $$$$\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)}={k} \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)}{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }={k} \\ $$$$\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }=\frac{{k}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }+\frac{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }=\frac{{k}}{\mathrm{2}}+\frac{\mathrm{2}}{{k}} \\ $$$$\frac{\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{4}} −{y}^{\mathrm{4}} \right)^{\mathrm{2}} }{\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{4}} −{y}^{\mathrm{4}} \right)}=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{k}} \\ $$$$\frac{\mathrm{2}\left({x}^{\mathrm{8}} +{y}^{\mathrm{8}} \right)}{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{2}{k}} \\ $$$$\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{k}} \\ $$$$\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }+\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }=\frac{{k}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{k}}+\frac{\mathrm{4}{k}}{{k}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({k}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{4}{k}\right)^{\mathrm{2}} }{\mathrm{4}{k}\left({k}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{k}^{\mathrm{4}} +\mathrm{24}{k}^{\mathrm{2}} +\mathrm{16}}{\mathrm{4}{k}\left({k}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$ \\ $$

Answered by bramlexs22 last updated on 04/Apr/21

The equality ((x^2 +y^2 )/(x^2 −y^2 ))+((x^2 −y^2 )/(x^2 +y^2 )) = k implies (((x^2 +y^2 )^2 +(x^2 −y^2 )^2 )/(x^4 −y^4 )) = k hence ((x^4 +y^4 )/(x^4 −y^4 )) = (k/2) .and ((x/y))^4 = ((k+2)/(k−2)) Therefore ((x^8 +y^8 )/(x^8 −y^8 )) + ((x^8 −y^8 )/(x^8 +y^8 )) = ((2(x^(16) +y^(16) ))/(x^(16) −y^(16) )) = 2((((((k+2)/(k−2)))^4 +1)/((((k−2)/(k+2)))^4 −1)))= 2((((k+2)^4 +(k−2)^4 )/((k+2)^4 −(k−2)^4 )))

$${The}\:{equality}\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:=\:{k} \\ $$$${implies}\:\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)^{\mathrm{2}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }\:=\:{k} \\ $$$${hence}\:\frac{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} }{{x}^{\mathrm{4}} −{y}^{\mathrm{4}} }\:=\:\frac{{k}}{\mathrm{2}}\:.{and}\:\left(\frac{{x}}{{y}}\right)^{\mathrm{4}} =\:\frac{{k}+\mathrm{2}}{{k}−\mathrm{2}}\: \\ $$$${Therefore} \\ $$$$\:\frac{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }\:+\:\frac{{x}^{\mathrm{8}} −{y}^{\mathrm{8}} }{{x}^{\mathrm{8}} +{y}^{\mathrm{8}} }\:=\:\frac{\mathrm{2}\left({x}^{\mathrm{16}} +{y}^{\mathrm{16}} \right)}{{x}^{\mathrm{16}} −{y}^{\mathrm{16}} } \\ $$$$=\:\mathrm{2}\left(\frac{\left(\frac{{k}+\mathrm{2}}{{k}−\mathrm{2}}\right)^{\mathrm{4}} +\mathrm{1}}{\left(\frac{{k}−\mathrm{2}}{{k}+\mathrm{2}}\right)^{\mathrm{4}} −\mathrm{1}}\right)=\:\mathrm{2}\left(\frac{\left({k}+\mathrm{2}\right)^{\mathrm{4}} +\left({k}−\mathrm{2}\right)^{\mathrm{4}} }{\left({k}+\mathrm{2}\right)^{\mathrm{4}} −\left({k}−\mathrm{2}\right)^{\mathrm{4}} }\right) \\ $$

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