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Question-4973




Question Number 4973 by hung260677 last updated on 28/Mar/16
Answered by Rasheed Soomro last updated on 30/Mar/16
((3x^2 +xy−y^2 )/(x+y))=9 ∧ ((2x^3 −x^2 y)/(x+y))=20  ⇒ x+y=((3x^2 +xy−y^2 )/9) ∧ x+y=((2x^3 −x^2 y)/(20))  ⇒ ((3x^2 +xy−y^2 )/9)=((2x^3 −x^2 y)/(20))  ⇒ 60x^2 +20xy−20y^2 =18x^3 −9x^2 y  ⇒18x^3 −60x^2 −9x^2 y−20xy+20y^2 =0  Any way of factorization?  Continue
$$\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{{x}+{y}}=\mathrm{9}\:\wedge\:\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{{x}+{y}}=\mathrm{20} \\ $$$$\Rightarrow\:{x}+{y}=\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{\mathrm{9}}\:\wedge\:{x}+{y}=\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{\mathrm{20}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{x}^{\mathrm{2}} +{xy}−{y}^{\mathrm{2}} }{\mathrm{9}}=\frac{\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} {y}}{\mathrm{20}} \\ $$$$\Rightarrow\:\mathrm{60}{x}^{\mathrm{2}} +\mathrm{20}{xy}−\mathrm{20}{y}^{\mathrm{2}} =\mathrm{18}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} {y} \\ $$$$\Rightarrow\mathrm{18}{x}^{\mathrm{3}} −\mathrm{60}{x}^{\mathrm{2}} −\mathrm{9}{x}^{\mathrm{2}} {y}−\mathrm{20}{xy}+\mathrm{20}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${Any}\:{way}\:{of}\:{factorization}? \\ $$$${Continue} \\ $$

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