Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 176472 by mr W last updated on 19/Sep/22

solve for x,y,z with x+y+z=x^2 +y^2 +z^2 =x^3 +y^3 +z^3 =5

$${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$

Answered by behi834171 last updated on 19/Sep/22

x+y+z=5 x^2 +x(y+z)=5x and so on ⇒Σx^2 +2Σxy=5Σx ⇒5+2Σxy=5×5⇒Σxy=10 Σx^2 =5⇒x^3 +x(y^2 +z^2 )=5x and so on.. ⇒Σx^3 +Σx(y^2 +z^2 )=5Σx⇒ ⇒Σx^3 +Σxy(x+y)=5Σx⇒ ⇒Σx^3 +Σxy(5−z)=5Σx⇒ ⇒Σx^3 +5Σxy−3xyz=5Σx ⇒5+5×10−3xyz=5×5⇒xyz=10 so: x,y,z; are the roots of : t^3 −5t^2 +10t−10=0 this have only one real root.

$${x}+{y}+{z}=\mathrm{5} \\ $$$${x}^{\mathrm{2}} +{x}\left({y}+{z}\right)=\mathrm{5}{x}\:\:{and}\:{so}\:{on} \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{2}} +\mathrm{2}\Sigma{xy}=\mathrm{5}\Sigma{x} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{2}\Sigma{xy}=\mathrm{5}×\mathrm{5}\Rightarrow\Sigma\boldsymbol{{xy}}=\mathrm{10} \\ $$$$\Sigma{x}^{\mathrm{2}} =\mathrm{5}\Rightarrow{x}^{\mathrm{3}} +{x}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{5}{x}\:\:{and}\:{so}\:{on}.. \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{x}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{xy}\left({x}+{y}\right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\Sigma{xy}\left(\mathrm{5}−{z}\right)=\mathrm{5}\Sigma{x}\Rightarrow \\ $$$$\Rightarrow\Sigma{x}^{\mathrm{3}} +\mathrm{5}\Sigma{xy}−\mathrm{3}{xyz}=\mathrm{5}\Sigma{x} \\ $$$$\Rightarrow\mathrm{5}+\mathrm{5}×\mathrm{10}−\mathrm{3}{xyz}=\mathrm{5}×\mathrm{5}\Rightarrow\boldsymbol{{xyz}}=\mathrm{10} \\ $$$$\boldsymbol{{so}}:\:\:{x},{y},{z};\:{are}\:{the}\:{roots}\:{of}\:: \\ $$$$\:\:\:\:\:\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{5}\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{{t}}−\mathrm{10}=\mathrm{0} \\ $$$${this}\:{have}\:{only}\:{one}\:{real}\:{root}. \\ $$

Commented by Tawa11 last updated on 20/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 19/Sep/22

thanks sir!

$${thanks}\:{sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com