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Question Number 211562 by MrGaster last updated on 13/Sep/24
{ ((x^2 +y^2 =1)),((x^3 −y=0)) :}
$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{1}}\\{\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{\mathrm{y}}=\mathrm{0}}\end{cases} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Sep/24
{ ((x^2 +y^2 =1...(i))),((x^3 −y=0...(ii))) :} (ii)⇒x^6 =y^2 (i)⇒ x^6 =1−x^2 x^6 +x^2 −1=0 (ii)⇒y=x^3 y^2 =x^6 y^2 =(x^2 )^3 y^2 =(1−y^2 )^3 y^2 =1−3y^2 +3y^4 −y^6 1−4y^2 +3y^4 −y^6 =0 ...
$$\:\:\:\:\:\:\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{1}…\left({i}\right)}\\{\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{\mathrm{y}}=\mathrm{0}…\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\Rightarrow{x}^{\mathrm{6}} ={y}^{\mathrm{2}} \\ $$$$\left({i}\right)\Rightarrow\:{x}^{\mathrm{6}} =\mathrm{1}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{6}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({ii}\right)\Rightarrow{y}={x}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} ={x}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\left(\mathrm{1}−{y}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\mathrm{1}−\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{4}} −{y}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{4}{y}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{4}} −{y}^{\mathrm{6}} =\mathrm{0} \\ $$$$… \\ $$$$ \\ $$
Answered by mr W last updated on 13/Sep/24
(x^2 )^3 +x^2 −1=0 x^2 =(((√((1/4)+(1/(27))))+(1/2)))^(1/3) −(((√((1/4)+(1/(27))))−(1/2)))^(1/3) ⇒x=±(√((((√((1/4)+(1/(27))))+(1/2)))^(1/3) −(((√((1/4)+(1/(27))))−(1/2)))^(1/3) ))
$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{x}=\pm\sqrt{\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}}+\frac{\mathrm{1}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$

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