Question Number 218041 by Mamadi last updated on 26/Mar/25
$${find}\:\left({x},{y},{z}\right)/{such}\:{as} \\ $$$$\left({x}+{y}+{z}=\mathrm{1}\right. \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:=\mathrm{9}\right. \\ $$$$\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\mathrm{1}\right. \\ $$
Answered by vnm last updated on 27/Mar/25
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{xz}+\mathrm{2}{yz}=\mathrm{1} \\ $$$${xy}+{xz}+{yz}=\frac{\mathrm{1}−\mathrm{9}}{\mathrm{2}}=−\mathrm{4} \\ $$$$\frac{{xy}+{xz}+{yz}}{{xyz}}=\mathrm{1} \\ $$$${xyz}=−\mathrm{4} \\ $$$$\left({t}−{x}\right)\left({t}−{y}\right)\left({t}−{z}\right)= \\ $$$${t}^{\mathrm{3}} −\left({x}+{y}+{z}\right){t}^{\mathrm{2}} +\left({xy}+{xz}+{yz}\right){t}−{xyz} \\ $$$${x},{y},{z}\:{are}\:{the}\:{roots}\:{of}\:{t}^{\mathrm{3}} −{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)−\mathrm{4}\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}+\mathrm{2}\right)\left({t}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left\{\mathrm{1},\:\mathrm{2},\:−\mathrm{2}\right\}\:{and}\:{all}\:{other}\:{permutations} \\ $$$${of}\:\:{these}\:{three}\:{numbers}. \\ $$