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