Answer to Question #191124 in Algebra for Yuki

Question

(x+y)³=(x-y)²

Answer 1

$Given,(x+y)³=(x-y)².\newline \text{Taking all terms to left side.}\newline (x+y)³-(x-y)²=0\newline \text{By using algebraic identities,}\newline (x+y)³=x^3+3x^2y+3xy^2+y^3 and (x-y)^2=x^2-2x^2y+y^2\newline Therefore,\newline (x+y)³-(x-y)²=0\newline (x^3+3x^2y+3xy^2+y^3)-(x^2-2xy+y^2)=0\newline x^3-x^2+3x^2y+3xy^2+2xy+y^3-y^2=0\newline (x^3+y^3)-(x^2+y^2)+3x^2y+3xy^2+2xy=0\newline (x+y)(x^2-xy+y^2)-(x^2+y^2)+3xy(x+y)+2xy=0\newline (x+y)(x^2-xy+y^2)-((x+y)^2-2xy)+3xy(x+y)+2xy=0\newline (x+y)(x^2-xy+y^2)-(x+y)^2+3xy(x+y)+4xy=0\newline (x+y)(x^2-xy+y^2-x-y+3xy)+4xy=0\newline (x+y)(x^2+2xy+y^2-x-y)+4xy=0\newline (x+y)((x+y)^2-(x+y))+4xy=0\newline (x+y)^2(x+y-1)+4xy=0\newline$