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Question-81857




Question Number 81857 by naka3546 last updated on 16/Feb/20
Commented by jagoll last updated on 16/Feb/20
zero = 0
$${zero}\:=\:\mathrm{0} \\ $$
Commented by john santu last updated on 16/Feb/20
x^2 +y^2 = ab(a+b)+(a+b)(a^2 −ab+b^2 )  x^2 +y^2 =(a+b)(a^2 +b^2 ) ∗  ⇒((x/a)−(y/b))((x/b)−(y/a)) =   ((ab(x^2 +y^2 )−(a^2 +b^2 )xy)/((ab)^2 )) =  ((ab(a+b)(a^2 +b^2 )−(ab)(a+b)(a^2 +b^2 ))/((ab)^2 )) =  0 .
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:{ab}\left({a}+{b}\right)+\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:\ast \\ $$$$\Rightarrow\left(\frac{{x}}{{a}}−\frac{{y}}{{b}}\right)\left(\frac{{x}}{{b}}−\frac{{y}}{{a}}\right)\:=\: \\ $$$$\frac{{ab}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){xy}}{\left({ab}\right)^{\mathrm{2}} }\:= \\ $$$$\frac{{ab}\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({ab}\right)\left({a}+{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{\left({ab}\right)^{\mathrm{2}} }\:= \\ $$$$\mathrm{0}\:.\: \\ $$

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