# x-2-y-2-xy-2-x-2-z-2-zx-1-y-2-z-2-yz-1-x-y-z-2-

Question Number 134410 by EDWIN88 last updated on 03/Mar/21
$$\:\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{2}}\\{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +{zx}=\mathrm{1}}\\{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{1}}\end{cases}\:\Rightarrow\:\left({x}+{y}+{z}\right)^{\mathrm{2}} \:=? \\$$
Commented by EDWIN88 last updated on 03/Mar/21
$$\mathrm{how}? \\$$
Commented by mr W last updated on 03/Mar/21
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{2}\pm\sqrt{\mathrm{3}} \\$$
Answered by mr W last updated on 03/Mar/21
$$\left({ii}\right)−\left({iii}\right): \\$$$$\left({x}−{y}\right)\left({x}+{y}\right)+{z}\left({x}−{y}\right)=\mathrm{0} \\$$$$\left({x}+{y}+{z}\right)\left({x}−{y}\right)=\mathrm{0} \\$$$$\Rightarrow{x}+{y}+{z}=\mathrm{0}\:…\left({I}\right)\:{or}\:{x}={y}\:\:\:…\left({II}\right) \\$$$$\\$$$$\left({i}\right)−\left({ii}\right): \\$$$$\left({y}−{z}\right)\left({y}+{z}\right)+{x}\left({y}−{z}\right)=\mathrm{1} \\$$$$\left({x}+{y}+{z}\right)\left({y}−{z}\right)=\mathrm{1}\:\:\:…\left({III}\right) \\$$$$\Rightarrow{x}+{y}+{z}\neq\mathrm{0},\:{y}\neq{z} \\$$$$\\$$$$\Rightarrow{x}={y} \\$$$$\left({i}\right):\:\mathrm{3}{x}^{\mathrm{2}} =\mathrm{2} \\$$$$\left({III}\right):\:{x}+{y}+{z}=\frac{\mathrm{1}}{{y}−{z}}=\frac{\mathrm{1}}{{x}−{z}} \\$$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +{xz}−\mathrm{3}{xz}}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\left(\mathrm{2}{x}+{z}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\mathrm{4}{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{4}{xz}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{xz}+{x}^{\mathrm{2}} +{z}^{\mathrm{2}} +{xz}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\mathrm{2}+\mathrm{3}{xz}+\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\mathrm{3}{xz}+\mathrm{3}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{3}{xz}} \\$$$$\mathrm{9}\left({xz}\right)^{\mathrm{2}} +\mathrm{6}{xz}−\mathrm{2}=\mathrm{0} \\$$$$\Rightarrow{xz}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{3}} \\$$$$\Rightarrow\mathrm{3}{xz}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\$$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}\pm\sqrt{\mathrm{3}}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\$$
Commented by EDWIN88 last updated on 03/Mar/21
$$\mathrm{nice} \\$$