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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
$(x^4 /q^3 ) + (y^4 /p^3 ) = 1 ⇒{((x^4 /q^3 ))+((y^4 /p^3 ))}×{((x^2 /q^3 ))+((y^2 /p^3 ))}= (x^6 /q^6 )+((x^4 y^2 )/(p^3 q^3 ))+((x^2 y^4 )/(p^3 q^3 ))+(y^6 /p^6 ) = (x^2 /q^3 )+(y^2 /p^3 ) (x^6 /q^6 ) +(y^6 /p^6 ) +((x^2 y^2 (x^2 +y^2 ))/(p^3 q^3 ))=((x^2 p^3 +y^2 q^3 )/(p^3 q^3 )) (x^6 /q^6 )+(y^6 /p^6 ) = ((x^2 p^3 +y^2 q^3 −x^2 y^2 )/(p^3 q^3 )) (1) (2)(x^2 +y^2 )×(p^3 +q^3 ) =1 ⇒x^2 p^3 +x^2 q^3 +y^2 p^3 +y^2 p^3 =1 x^2 p^3 +y^2 q^3 = 1−x^2 q^3 −y^2 p^3 substitute (2) in (1) ⇒ (x^6 /q^6 )+(y^6 /p^6 ) =((1−x^2 q^3 −y^2 p^3 −x^2 y^2 )/(p^3 q^3 ))$
$\frac{x^{4}}{q^{3}} + \frac{y^{4}}{p^{3}} = 1$ $\Rightarrow {(\frac{x^{4}}{q^{3}}) + (\frac{y^{4}}{p^{3}})} \times {(\frac{x^{2}}{q^{3}}) + (\frac{y^{2}}{p^{3}})} =$ $\frac{x^{6}}{q^{6}} + \frac{x^{4} y^{2}}{p^{3} q^{3}} + \frac{x^{2} y^{4}}{p^{3} q^{3}} + \frac{y^{6}}{p^{6}} = \frac{x^{2}}{q^{3}} + \frac{y^{2}}{p^{3}}$ $\frac{x^{6}}{q^{6}} + \frac{y^{6}}{p^{6}} + \frac{x^{2} y^{2} (x^{2} + y^{2})}{p^{3} q^{3}} = \frac{x^{2} p^{3} + y^{2} q^{3}}{p^{3} q^{3}}$ $\frac{x^{6}}{q^{6}} + \frac{y^{6}}{p^{6}} = \frac{x^{2} p^{3} + y^{2} q^{3} - x^{2} y^{2}}{p^{3} q^{3}} (1)$ $(2) (x^{2} + y^{2}) \times (p^{3} + q^{3}) = 1$ $\Rightarrow x^{2} p^{3} + x^{2} q^{3} + y^{2} p^{3} + y^{2} p^{3} = 1$ $x^{2} p^{3} + y^{2} q^{3} = 1 - x^{2} q^{3} - y^{2} p^{3}$ $s u b s t i t u t e (2) i n (1)$ $\Rightarrow \frac{x^{6}}{q^{6}} + \frac{y^{6}}{p^{6}} = \frac{1 - x^{2} q^{3} - y^{2} p^{3} - x^{2} y^{2}}{p^{3} q^{3}}$
Commented by TawaTawa last updated on 08/Feb/20

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