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

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