Question Number 159179 by cortano last updated on 14/Nov/21
$$\:\:\mathcal{X}^{\mathrm{3}} −\mathrm{4}\mathcal{X}^{\mathrm{2}} −\mathrm{6}\mathcal{X}−\mathrm{24}\:=\:\mathrm{0} \\ $$
Answered by mr W last updated on 14/Nov/21
$${let}\:\mathcal{X}={x}+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}×\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{3}×\frac{\mathrm{16}}{\mathrm{9}}{x}+\frac{\mathrm{64}}{\mathrm{27}}−\mathrm{4}×{x}^{\mathrm{2}} −\mathrm{4}×\mathrm{2}{x}×\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{4}×\frac{\mathrm{16}}{\mathrm{9}}−\mathrm{6}{x}−\mathrm{6}×\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{24}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{34}{x}}{\mathrm{3}}−\frac{\mathrm{992}}{\mathrm{27}}=\mathrm{0} \\ $$$$\Delta=\sqrt{−\frac{\mathrm{34}^{\mathrm{3}} }{\mathrm{9}^{\mathrm{3}} }+\frac{\mathrm{496}^{\mathrm{2}} }{\mathrm{27}^{\mathrm{2}} }}=\frac{\mathrm{2}\sqrt{\mathrm{638}}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{638}}}{\mathrm{3}}+\frac{\mathrm{496}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{638}}}{\mathrm{3}}−\frac{\mathrm{496}}{\mathrm{27}}} \\ $$$$\Rightarrow\mathcal{X}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}+\sqrt[{\mathrm{3}}]{\mathrm{18}\sqrt{\mathrm{638}}+\mathrm{496}}−\sqrt[{\mathrm{3}}]{\mathrm{18}\sqrt{\mathrm{638}}−\mathrm{496}}\right)\approx\mathrm{5}.\mathrm{7635} \\ $$
Commented by cortano last updated on 14/Nov/21
$${waw}\:{amazing} \\ $$