Question Number 158855 by LEKOUMA last updated on 09/Nov/21
$${Resolve}\:{the}\:{system}\:{d}'\:{unknow}\:\:\left({x},\:{y},{z}\right)\:\in\:\subset^{\mathrm{3}} \\ $$$${x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =−\mathrm{5} \\ $$
Answered by mr W last updated on 09/Nov/21
$${x}+{y}+{z}=\mathrm{1} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{xy}+{yz}+{zx}=\mathrm{0} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)−\mathrm{3}{xyz} \\ $$$$\mathrm{1}^{\mathrm{3}} =−\mathrm{5}+\mathrm{3}×\mathrm{1}×\mathrm{0}−\mathrm{3}{xyz} \\ $$$$\Rightarrow{xyz}=−\mathrm{2} \\ $$$${x},{y},{z}\:{are}\:{roots}\:{of} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}−\mathrm{2}{t}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(−\mathrm{1},\mathrm{1}+{i},\mathrm{1}−{i}\right) \\ $$
Commented by LEKOUMA last updated on 09/Nov/21
$${Many}\:{thanks} \\ $$