Question Number 176472 by mr W last updated on 19/Sep/22
$${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$
Answered by behi834171 last updated on 19/Sep/22
$${x}+{y}+{z}=\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{x}\left({y}+{z}\right)=\mathrm{5}{x}\:\:{and}\:{so}\:{on} \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{2}} +\mathrm{2}\Sigma{xy}=\mathrm{5}\Sigma{x} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{2}\Sigma{xy}=\mathrm{5}×\mathrm{5}\Rightarrow\Sigma\boldsymbol{{xy}}=\mathrm{10} \\ $$$$\Sigma{x}^{\mathrm{2}} =\mathrm{5}\Rightarrow{x}^{\mathrm{3}} +{x}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{5}{x}\:\:{and}\:{so}\:{on}.. \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{x}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{xy}\left({x}+{y}\right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{xy}\left(\mathrm{5}−{z}\right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\mathrm{5}\Sigma{xy}−\mathrm{3}{xyz}=\mathrm{5}\Sigma{x} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{5}×\mathrm{10}−\mathrm{3}{xyz}=\mathrm{5}×\mathrm{5}\Rightarrow\boldsymbol{{xyz}}=\mathrm{10} \\ $$$$\boldsymbol{{so}}:\:\:{x},{y},{z};\:{are}\:{the}\:{roots}\:{of}\:: \\ $$$$\:\:\:\:\:\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{5}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{{t}}−\mathrm{10}=\mathrm{0} \\ $$$${this}\:{have}\:{only}\:{one}\:{real}\:{root}. \\ $$
Commented by Tawa11 last updated on 20/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 19/Sep/22
$${thanks}\:{sir}! \\ $$