Question-155928

Question Number 155928 by cortano last updated on 06/Oct/21

Answered by mr W last updated on 06/Oct/21

(u+v)^3 =u^3 +v^3 +3uv(u+v) (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0 x^3 −3x+4=0 ⇒x=u+v ⇒uv=1 ⇒u^3 v^3 =1 ⇒u^3 +v^3 =−4 u^3 ,v^3 are roots of z^2 +4z+1=0 ⇒u^3 ,v^3 =z=−2±(√3) ⇒u=((−2+(√3)))^(1/3) , v=((−2−(√3)))^(1/3) ⇒x=u+v=((−2+(√3)))^(1/3) +((−2−(√3)))^(1/3) ⇒x=−((2−(√3)))^(1/3) −((2+(√3)))^(1/3)

$$\left({u}+{v}\right)^{\mathrm{3}} ={u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}{uv}\left({u}+{v}\right) \\ $$$$\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}={u}+{v} \\ $$$$\Rightarrow{uv}=\mathrm{1}\:\Rightarrow{u}^{\mathrm{3}} {v}^{\mathrm{3}} =\mathrm{1} \\ $$$$\Rightarrow{u}^{\mathrm{3}} +{v}^{\mathrm{3}} =−\mathrm{4} \\ $$$${u}^{\mathrm{3}} ,{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{3}} ,{v}^{\mathrm{3}} ={z}=−\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{u}=\sqrt[{\mathrm{3}}]{−\mathrm{2}+\sqrt{\mathrm{3}}},\:{v}=\sqrt[{\mathrm{3}}]{−\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}={u}+{v}=\sqrt[{\mathrm{3}}]{−\mathrm{2}+\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{3}}]{−\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{3}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$

Commented by cortano last updated on 06/Oct/21