Question Number 146158 by aliibrahim1 last updated on 11/Jul/21
Commented by MJS_new last updated on 12/Jul/21
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z} \\ $$$${z}=\mathrm{10}\vee{z}=\mathrm{10}\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$
Answered by mnjuly1970 last updated on 11/Jul/21
$$\:\:{i}\::\:\:\:{x}^{\:\mathrm{6}} −\mathrm{6}{x}^{\:\mathrm{4}} {y}^{\:\mathrm{2}} +\mathrm{9}{x}^{\mathrm{2}} {y}^{\:\mathrm{4}} =\mathrm{900}\:\: \\ $$$$\:\:{ii}\::\:\:{y}^{\:\mathrm{6}} −\mathrm{6}{y}^{\:\mathrm{4}} {x}^{\:\mathrm{2}} +\:\mathrm{9}{y}^{\:\mathrm{2}} {x}^{\mathrm{4}} =\mathrm{100} \\ $$$$\:{i}\:+\:{ii}\::\:\:{x}^{\:\mathrm{6}} +\mathrm{3}{x}^{\:\mathrm{2}} {y}^{\:\mathrm{4}} +\mathrm{3}{x}^{\:\mathrm{4}} {y}^{\:\mathrm{2}} +{y}^{\:\mathrm{6}} =\:\mathrm{1000} \\ $$$$\:\:\:\left({x}^{\:\mathrm{2}} \:+\:{y}^{\:\mathrm{2}} \right)^{\:\mathrm{3}} =\:\mathrm{1000} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\:\mathrm{2}} \:=\mathrm{10}\:.... \\ $$
Commented by aliibrahim1 last updated on 11/Jul/21
$${thx}\:{sir} \\ $$
Commented by mnjuly1970 last updated on 11/Jul/21
$${you}\:{are}\:{welcom}\:... \\ $$