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Question Number 177928 by Spillover last updated on 11/Oct/22

Prove that  [((x^3 +y^3 )/((x−y)^2 +3y))]÷[(((x+y)^2 −3xy)/(x^3 −y^3 ))]×[((xy)/(x^2 −y^2 ))]=xy

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} }{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{3y}}\right]\boldsymbol{\div}\left[\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{3xy}}{\mathrm{x}^{\mathrm{3}} −\mathrm{y}^{\mathrm{3}} }\right]×\left[\frac{\mathrm{xy}}{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right]=\mathrm{xy} \\ $$

Answered by Ar Brandon last updated on 11/Oct/22

((x^3 +y^3 )/((x−y)^2 +3yx))∙((x^3 −y^3 )/((x+y)^2 −3xy))∙((xy)/(x^2 −y^2 ))  =((x^3 +y^3 )/(x^2 +xy+y^2 ))∙((x^3 −y^3 )/(x^2 −xy+y^2 ))∙((xy)/((x−y)(x+y)))  =(((x^3 +y^3 )(x−y))/(x^3 −y^3 ))∙(((x^3 −y^3 )(x+y))/(x^3 +y^3 ))∙((xy)/((x−y)(x+y)))  =(x−y)(x+y)∙((xy)/((x−y)(x+y)))=xy

$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{3}{yx}}\centerdot\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}{xy}}\centerdot\frac{{xy}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} }\centerdot\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} }\centerdot\frac{{xy}}{\left({x}−{y}\right)\left({x}+{y}\right)} \\ $$$$=\frac{\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)\left({x}−{y}\right)}{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }\centerdot\frac{\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right)\left({x}+{y}\right)}{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }\centerdot\frac{{xy}}{\left({x}−{y}\right)\left({x}+{y}\right)} \\ $$$$=\left({x}−{y}\right)\left({x}+{y}\right)\centerdot\frac{{xy}}{\left({x}−{y}\right)\left({x}+{y}\right)}={xy} \\ $$

Commented by Spillover last updated on 11/Oct/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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