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Question Number 62234 by aliesam last updated on 18/Jun/19

Answered by tanmay last updated on 18/Jun/19

9(x+y)=(x+y)(x^2 −xy+y^2 ) (x+y)(9−x^2 +xy−y^2 )=0 eithdr x+y=0 or x^2 −xy+y^2 =9 x=−y y^3 =−2y+7y y(y^2 −5)=0 either y=0 or y^2 =5 so x=0 y=0 or x=−(√( 5)) y=(√5) and x=(√5) y=−(√5) 5(x−y)=(x−y)(x^2 +xy+y^2 ) (x−y)=0 or x^2 +xy+y^2 =5 x^3 =7x+2x x(x^2 −9)=0 whenx^2 −9=0 x=±3 so y=±3

9(x+y)=(x+y)(x2−xy+y2)(x+y)(9−x2+xy−y2)=0eithdrx+y=0orx2−xy+y2=9x=−yy3=−2y+7yy(y2−5)=0eithery=0ory2=5sox=0y=0orx=−5y=5andx=5y=−55(x−y)=(x−y)(x2+xy+y2)(x−y)=0orx2+xy+y2=5x3=7x+2xx(x2−9)=0whenx2−9=0x=±3soy=±3

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