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Question Number 211562 by MrGaster last updated on 13/Sep/24

$$$$$$\{\begin{array}{l}{\mathit{x}}^2+{\mathbf{y}}^2=1\\ {\mathit{x}}^3-\mathbf{y}=0\end{array}$$

Answered by Rasheed.Sindhi last updated on 13/Sep/24

$$\{\begin{array}{l}{\mathit{x}}^2+{\mathbf{y}}^2=1...\left(i\right)\\ {\mathit{x}}^3-\mathbf{y}=0...\left(ii\right)\end{array}$$$$\left(ii\right)\Rightarrow x^6=y^2$$$$\left(i\right)\Rightarrow x^6=1-x^2$$$$x^6+x^2-1=0$$$$\left(ii\right)\Rightarrow y=x^3$$$$y^2=x^6$$$$y^2={\left(x^2\right)}^3$$$$y^2={(1-y^2)}^3$$$$y^2=1-3y^2+3y^4-y^6$$$$1-4y^2+3y^4-y^6=0$$$$...$$$$$$

Answered by mr W last updated on 13/Sep/24

$${\left(x^2\right)}^3+x^2-1=0$$$$x^2=\sqrt[3]{\sqrt{\frac{1}{4}+\frac{1}{27}}+\frac{1}{2}}-\sqrt[3]{\sqrt{\frac{1}{4}+\frac{1}{27}}-\frac{1}{2}}$$$$\Rightarrow x=\pm \sqrt{\sqrt[3]{\sqrt{\frac{1}{4}+\frac{1}{27}}+\frac{1}{2}}-\sqrt[3]{\sqrt{\frac{1}{4}+\frac{1}{27}}-\frac{1}{2}}}$$

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