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Question Number 190942 by mathlove last updated on 14/Apr/23

(x+y)(x^2 −xy+y^2 )=2  ((x^3 y^3 )/(x^9 +y^9 −8))=?

$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\frac{{x}^{\mathrm{3}} {y}^{\mathrm{3}} }{{x}^{\mathrm{9}} +{y}^{\mathrm{9}} −\mathrm{8}}=? \\ $$

Answered by Frix last updated on 15/Apr/23

x^3 +y^3 =2 ⇒ y^3 =2−x^3   ((x^3 (2−x^3 ))/(x^9 +(2−x^3 )^3 −8))=((x^3 (2−x^3 ))/(−6x^3 (2−x^3 )))=−(1/6)

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{2}\:\Rightarrow\:{y}^{\mathrm{3}} =\mathrm{2}−{x}^{\mathrm{3}} \\ $$$$\frac{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{9}} +\left(\mathrm{2}−{x}^{\mathrm{3}} \right)^{\mathrm{3}} −\mathrm{8}}=\frac{{x}^{\mathrm{3}} \left(\mathrm{2}−{x}^{\mathrm{3}} \right)}{−\mathrm{6}{x}^{\mathrm{3}} \left(\mathrm{2}−{x}^{\mathrm{3}} \right)}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by manxsol last updated on 15/Apr/23

x^3 +y^3 =2  (x^3 +y^3 )^3 =x^9 +y^9 +3x^3 y^3 (x^3 +y^3 )  8=x^9 +y^9 +6x^3 y^3   −6x^3 y^3 =x^9 +y^9 −8  ((x^3 y^3 )/(−6x^3 y^3 ))=−(1/6)

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{2} \\ $$$$\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)^{\mathrm{3}} ={x}^{\mathrm{9}} +{y}^{\mathrm{9}} +\mathrm{3}{x}^{\mathrm{3}} {y}^{\mathrm{3}} \left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right) \\ $$$$\mathrm{8}={x}^{\mathrm{9}} +{y}^{\mathrm{9}} +\mathrm{6}{x}^{\mathrm{3}} {y}^{\mathrm{3}} \\ $$$$−\mathrm{6}{x}^{\mathrm{3}} {y}^{\mathrm{3}} ={x}^{\mathrm{9}} +{y}^{\mathrm{9}} −\mathrm{8} \\ $$$$\frac{{x}^{\mathrm{3}} {y}^{\mathrm{3}} }{−\mathrm{6}{x}^{\mathrm{3}} {y}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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