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Question Number 110815 by I want to learn more last updated on 30/Aug/20

Answered by mr W last updated on 30/Aug/20

$${(x+y)}^3=x^3+y^3+3xy(x+y)$$$${(x+y)}^3={(x+y)}^2+3xy(x+y)$$$$(x+y)[{(x+y)}^2-(x+y)-3xy]=0$$$$\Rightarrow x+y=0\Rightarrow x=-y=k$$$$\Rightarrow {(x+y)}^2-(x+y)-3xy=0$$$$\Rightarrow (x+y)(x+y-1)=3xy$$$$\Rightarrow x+y=3\Rightarrow 2=xy\Rightarrow x=2,y=1or$$$$x=1,y=2$$$$\Rightarrow x+y-1=3\Rightarrow 4=xy\Rightarrow x=y=2$$$$\Rightarrow x+y-1=0\Rightarrow x=0,y=1orx=1,y=0$$$$$$$$allsolutions:$$$$x=k,y=-kwithk\in \mathbb{Z}$$$$x=1,y=2$$$$x=2,y=1$$$$x=2,y=2$$$$x=0,y=1$$$$x=1,y=0$$

Commented by I want to learn more last updated on 31/Aug/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

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