Question Number 200961 by jabarsing last updated on 27/Nov/23
$$\: \\ $$$$\:\:\:\:\:−{x}^{\mathrm{3}} +\mathrm{1}=\sqrt[{\mathrm{3}}]{−{x}+\mathrm{1}}\:\Rightarrow{x}=? \\ $$$$ \\ $$
Answered by cortano12 last updated on 27/Nov/23
Answered by Frix last updated on 27/Nov/23
$$\mathrm{1}−{x}^{\mathrm{3}} =\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{3}} \right)^{\mathrm{3}} =\mathrm{1}−{x} \\ $$$${x}^{\mathrm{9}} −\mathrm{3}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{3}} −{x}=\mathrm{0} \\ $$$${x}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{93}}}{\mathrm{18}}}\approx.\mathrm{682327804} \\ $$$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}−\sqrt{−\mathrm{2}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{4}}\approx−.\mathrm{535687387} \\ $$$${x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}+\sqrt{−\mathrm{2}+\mathrm{6}\sqrt{\mathrm{5}}}}{\mathrm{4}}\approx\mathrm{1}.\mathrm{15372138} \\ $$