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 ⇒x+y=0 ⇒x=−y=k ⇒(x+y)^2 −(x+y)−3xy=0 ⇒(x+y)(x+y−1)=3xy ⇒x+y=3 ⇒2=xy⇒x=2,y=1 or x=1,y=2 ⇒x+y−1=3⇒4=xy⇒x=y=2 ⇒x+y−1=0⇒x=0, y=1 or x=1, y=0 all solutions: x=k, y=−k with k∈Z x=1, y=2 x=2, y=1 x=2, y=2 x=0, y=1 x=1, y=0](https://www.tinkutara.com/question/Q110821.png)
$$\left({x}+{y}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} =\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$\left({x}+{y}\right)\left[\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)−\mathrm{3}{xy}\right]=\mathrm{0} \\ $$$$\Rightarrow{x}+{y}=\mathrm{0}\:\Rightarrow{x}=−{y}={k} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)−\mathrm{3}{xy}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+{y}\right)\left({x}+{y}−\mathrm{1}\right)=\mathrm{3}{xy} \\ $$$$\Rightarrow{x}+{y}=\mathrm{3}\:\Rightarrow\mathrm{2}={xy}\Rightarrow{x}=\mathrm{2},{y}=\mathrm{1}\:{or} \\ $$$${x}=\mathrm{1},{y}=\mathrm{2} \\ $$$$\Rightarrow{x}+{y}−\mathrm{1}=\mathrm{3}\Rightarrow\mathrm{4}={xy}\Rightarrow{x}={y}=\mathrm{2} \\ $$$$\Rightarrow{x}+{y}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{0},\:{y}=\mathrm{1}\:{or}\:{x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$$ \\ $$$${all}\:{solutions}: \\ $$$${x}={k},\:{y}=−{k}\:{with}\:{k}\in\mathbb{Z} \\ $$$${x}=\mathrm{1},\:{y}=\mathrm{2} \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{1} \\ $$$${x}=\mathrm{2},\:{y}=\mathrm{2} \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$
Commented by I want to learn more last updated on 31/Aug/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$