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Question Number 202082 by Abduljalal last updated on 19/Dec/23

Answered by deleteduser1 last updated on 20/Dec/23

$$p^2-p-2=0(p=x^3)\Rightarrow p=2or-1$$$$\Rightarrow x^3=2or{x}^3=-1\Rightarrow x=\sqrt[3]{2};\sqrt[3]{2}{e}^{\frac{2i\pi }{3}},\sqrt[3]{2}{e}^{\frac{4\pi i}{3}}$$$$x=e^{\frac{\pi i}{3}};-e^{\frac{2\pi i}{3}};-1$$

Answered by a.lgnaoui last updated on 20/Dec/23

$${\mathbf{x}}^3({\mathbf{x}}^3-1)=2\mathbf{posons}{\mathbf{x}}^3=\mathbf{X}$$$$\mathbf{X}(\mathbf{X}-1)=2$$$${\mathbf{X}}^2-\mathbf{X}-2=0\mathbf{X}=\frac{1\pm 3}{2}$$$$\mathbf{X}=\{\begin{array}{l}-1\\ +2\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathbf{x}={(-1)}^{1/3}\Rightarrow \{\begin{array}{l}\mathbf{x}=-1\\ \mathbf{x}=\{\frac{1+\mathrm{i}\sqrt{3}}{2},\frac{1-\mathrm{i}\sqrt{3}}{2}\end{array}\\ \mathbf{x}=2^{1/3}\end{array}$$$$\mathbf{S}=\{-1,\frac{1-\mathrm{i}\sqrt{3}}{2};\frac{1+\mathrm{i}\sqrt{3}}{2};{}^3\sqrt{2}\}$$

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