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Question Number 80912 by TawaTawa last updated on 07/Feb/20

Commented by john santu last updated on 08/Feb/20

(x^4 /q^3 ) + (y^4 /p^3 ) = 1 ⇒{((x^4 /q^3 ))+((y^4 /p^3 ))}×{((x^2 /q^3 ))+((y^2 /p^3 ))}= (x^6 /q^6 )+((x^4 y^2 )/(p^3 q^3 ))+((x^2 y^4 )/(p^3 q^3 ))+(y^6 /p^6 ) = (x^2 /q^3 )+(y^2 /p^3 ) (x^6 /q^6 ) +(y^6 /p^6 ) +((x^2 y^2 (x^2 +y^2 ))/(p^3 q^3 ))=((x^2 p^3 +y^2 q^3 )/(p^3 q^3 )) (x^6 /q^6 )+(y^6 /p^6 ) = ((x^2 p^3 +y^2 q^3 −x^2 y^2 )/(p^3 q^3 )) (1) (2)(x^2 +y^2 )×(p^3 +q^3 ) =1 ⇒x^2 p^3 +x^2 q^3 +y^2 p^3 +y^2 p^3 =1 x^2 p^3 +y^2 q^3 = 1−x^2 q^3 −y^2 p^3 substitute (2) in (1) ⇒ (x^6 /q^6 )+(y^6 /p^6 ) =((1−x^2 q^3 −y^2 p^3 −x^2 y^2 )/(p^3 q^3 ))

$$\frac{{x}^{\mathrm{4}} }{{q}^{\mathrm{3}} }\:+\:\frac{{y}^{\mathrm{4}} }{{p}^{\mathrm{3}} }\:=\:\mathrm{1}\: \\ $$$$\Rightarrow\left\{\left(\frac{{x}^{\mathrm{4}} }{{q}^{\mathrm{3}} }\right)+\left(\frac{{y}^{\mathrm{4}} }{{p}^{\mathrm{3}} }\right)\right\}×\left\{\left(\frac{{x}^{\mathrm{2}} }{{q}^{\mathrm{3}} }\right)+\left(\frac{{y}^{\mathrm{2}} }{{p}^{\mathrm{3}} }\right)\right\}= \\ $$$$\frac{{x}^{\mathrm{6}} }{{q}^{\mathrm{6}} }+\frac{{x}^{\mathrm{4}} {y}^{\mathrm{2}} }{{p}^{\mathrm{3}} {q}^{\mathrm{3}} }+\frac{{x}^{\mathrm{2}} {y}^{\mathrm{4}} }{{p}^{\mathrm{3}} {q}^{\mathrm{3}} }+\frac{{y}^{\mathrm{6}} }{{p}^{\mathrm{6}} }\:=\:\frac{{x}^{\mathrm{2}} }{{q}^{\mathrm{3}} }+\frac{{y}^{\mathrm{2}} }{{p}^{\mathrm{3}} } \\ $$$$\frac{{x}^{\mathrm{6}} }{{q}^{\mathrm{6}} }\:+\frac{{y}^{\mathrm{6}} }{{p}^{\mathrm{6}} }\:+\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{{p}^{\mathrm{3}} {q}^{\mathrm{3}} }=\frac{{x}^{\mathrm{2}} {p}^{\mathrm{3}} +{y}^{\mathrm{2}} {q}^{\mathrm{3}} }{{p}^{\mathrm{3}} {q}^{\mathrm{3}} } \\ $$$$\frac{{x}^{\mathrm{6}} }{{q}^{\mathrm{6}} }+\frac{{y}^{\mathrm{6}} }{{p}^{\mathrm{6}} }\:=\:\:\frac{{x}^{\mathrm{2}} {p}^{\mathrm{3}} +{y}^{\mathrm{2}} {q}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{p}^{\mathrm{3}} {q}^{\mathrm{3}} }\:\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)×\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)\:=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {p}^{\mathrm{3}} +{x}^{\mathrm{2}} {q}^{\mathrm{3}} +{y}^{\mathrm{2}} {p}^{\mathrm{3}} +{y}^{\mathrm{2}} {p}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{2}} {p}^{\mathrm{3}} +{y}^{\mathrm{2}} {q}^{\mathrm{3}} \:=\:\mathrm{1}−{x}^{\mathrm{2}} {q}^{\mathrm{3}} −{y}^{\mathrm{2}} {p}^{\mathrm{3}} \\ $$$${substitute}\:\left(\mathrm{2}\right)\:{in}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\frac{{x}^{\mathrm{6}} }{{q}^{\mathrm{6}} }+\frac{{y}^{\mathrm{6}} }{{p}^{\mathrm{6}} }\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} {q}^{\mathrm{3}} −{y}^{\mathrm{2}} {p}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{p}^{\mathrm{3}} {q}^{\mathrm{3}} } \\ $$$$ \\ $$

Commented by TawaTawa last updated on 08/Feb/20

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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