Question Number 202082 by Abduljalal last updated on 19/Dec/23
Answered by AST last updated on 20/Dec/23
$${p}^{\mathrm{2}} −{p}−\mathrm{2}=\mathrm{0}\left({p}={x}^{\mathrm{3}} \right)\Rightarrow{p}=\mathrm{2}\:{or}\:−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\mathrm{2}\:{or}\:{x}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}};\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{3}}} ,\sqrt[{\mathrm{3}}]{\mathrm{2}}{e}^{\frac{\mathrm{4}\pi{i}}{\mathrm{3}}} \\ $$$${x}={e}^{\frac{\pi{i}}{\mathrm{3}}} ;−{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} ;−\mathrm{1} \\ $$
Answered by a.lgnaoui last updated on 20/Dec/23
$$\boldsymbol{\mathrm{x}}^{\mathrm{3}} \left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{2}\:\:\:\boldsymbol{\mathrm{posons}}\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} =\boldsymbol{\mathrm{X}} \\ $$$$\boldsymbol{\mathrm{X}}\left(\boldsymbol{\mathrm{X}}−\mathrm{1}\right)=\mathrm{2} \\ $$$$\boldsymbol{\mathrm{X}}^{\mathrm{2}} −\boldsymbol{\mathrm{X}}−\mathrm{2}=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{X}}=\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{X}}=\begin{cases}{−\mathrm{1}}\\{+\mathrm{2}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{\mathrm{x}}=\left(−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \Rightarrow\begin{cases}{\boldsymbol{\mathrm{x}}=−\mathrm{1}}\\{\boldsymbol{\mathrm{x}}=\left\{\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}},\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right.}\end{cases}}\\{\boldsymbol{\mathrm{x}}=\mathrm{2}^{\mathrm{1}/\mathrm{3}} }\end{cases} \\ $$$$\:\:\boldsymbol{\mathrm{S}}=\left\{−\mathrm{1},\:\:\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}};\:\:\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}};\:\:\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right\} \\ $$