Question #191124
(x+y)³=(x-y




1
Expert's answer
2021-05-10T17:02:20-0400

Given,(x+y)3=(xy)2.Taking all terms to left side.(x+y)3(xy)2=0By using algebraic identities,(x+y)3=x3+3x2y+3xy2+y3and(xy)2=x22x2y+y2Therefore,(x+y)3(xy)2=0(x3+3x2y+3xy2+y3)(x22xy+y2)=0x3x2+3x2y+3xy2+2xy+y3y2=0(x3+y3)(x2+y2)+3x2y+3xy2+2xy=0(x+y)(x2xy+y2)(x2+y2)+3xy(x+y)+2xy=0(x+y)(x2xy+y2)((x+y)22xy)+3xy(x+y)+2xy=0(x+y)(x2xy+y2)(x+y)2+3xy(x+y)+4xy=0(x+y)(x2xy+y2xy+3xy)+4xy=0(x+y)(x2+2xy+y2xy)+4xy=0(x+y)((x+y)2(x+y))+4xy=0(x+y)2(x+y1)+4xy=0Given,(x+y)³=(x-y)².\newline \text{Taking all terms to left side.}\newline (x+y)³-(x-y)²=0\newline \text{By using algebraic identities,}\newline (x+y)³=x^3+3x^2y+3xy^2+y^3 and (x-y)^2=x^2-2x^2y+y^2\newline Therefore,\newline (x+y)³-(x-y)²=0\newline (x^3+3x^2y+3xy^2+y^3)-(x^2-2xy+y^2)=0\newline x^3-x^2+3x^2y+3xy^2+2xy+y^3-y^2=0\newline (x^3+y^3)-(x^2+y^2)+3x^2y+3xy^2+2xy=0\newline (x+y)(x^2-xy+y^2)-(x^2+y^2)+3xy(x+y)+2xy=0\newline (x+y)(x^2-xy+y^2)-((x+y)^2-2xy)+3xy(x+y)+2xy=0\newline (x+y)(x^2-xy+y^2)-(x+y)^2+3xy(x+y)+4xy=0\newline (x+y)(x^2-xy+y^2-x-y+3xy)+4xy=0\newline (x+y)(x^2+2xy+y^2-x-y)+4xy=0\newline (x+y)((x+y)^2-(x+y))+4xy=0\newline (x+y)^2(x+y-1)+4xy=0\newline


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