Question Number 122783 by mathocean1 last updated on 19/Nov/20
$${solve}\:: \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Nov/20
$${x}={y}−\mathrm{1} \\ $$$$\left({y}−\mathrm{1}\right)^{\mathrm{3}} +\mathrm{3}\left({y}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}−\mathrm{1}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{3}−\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}=−\mathrm{1}\:\:\:\:\:\:\: \\ $$$${y}={s}−{t}\:\:\:\:\:\: \\ $$$$\left({s}−{t}\right)^{\mathrm{3}} −\mathrm{3}\left({s}−{t}\right)=−\mathrm{1} \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} −\mathrm{3}{st}\left({s}−{t}\right)−\mathrm{3}\left({s}−{t}\right)=−\mathrm{1} \\ $$$${st}=−\mathrm{1} \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} =−\mathrm{1}\Rightarrow−\frac{\mathrm{1}}{{t}^{\mathrm{3}} }−{t}^{\mathrm{3}} =−\mathrm{1}\Rightarrow{t}^{\mathrm{6}} −{t}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\Rightarrow{t}^{\mathrm{3}} =\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$${s}=\sqrt[{\mathrm{3}}]{\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}}−\mathrm{1} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Nov/20
$${Generally} \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${then}\:{depressed}\:{equation} \\ $$$${substitute}\:{x}={y}−\frac{{b}}{\mathrm{3}{a}} \\ $$$${a}\left({y}−\frac{{b}}{\mathrm{3}{a}}\right)^{\mathrm{3}} +{b}\left({y}−\frac{{b}}{\mathrm{3}{a}}\right)^{\mathrm{2}} +{c}\left({y}−\frac{{b}}{\mathrm{3}{a}}\right)+{d}=\mathrm{0} \\ $$$${ay}^{\mathrm{3}} −{by}\left({y}−\frac{{b}}{\mathrm{3}{a}}\right)−\frac{{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{2}} }+{by}^{\mathrm{2}} −\frac{\mathrm{2}{b}^{\mathrm{2}} {y}}{\mathrm{3}{a}}+\frac{{b}^{\mathrm{3}} }{\mathrm{9}{a}^{\mathrm{2}} }+{cy}−\frac{{bc}}{\mathrm{3}{a}}+{d}=\mathrm{0} \\ $$$${ay}^{\mathrm{3}} +\frac{{b}^{\mathrm{2}} {y}}{\mathrm{3}{a}}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{b}^{\mathrm{2}} {y}+{bc}}{\mathrm{3}{a}}+{d}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} −{y}\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }+\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{d}}{{a}}\right)=\mathrm{0} \\ $$$${y}={s}−{t}\:\: \\ $$$$\mathrm{3}{st}=\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }\:\Rightarrow{t}=\frac{{b}^{\mathrm{2}} }{\mathrm{9}{a}^{\mathrm{2}} {s}} \\ $$$${s}^{\mathrm{3}} −{t}^{\mathrm{3}} =−\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{d}}{{a}}\right) \\ $$$${s}^{\mathrm{3}} −\frac{{b}^{\mathrm{6}} }{\mathrm{729}{a}^{\mathrm{6}} {s}^{\mathrm{3}} }=\left(\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{d}}{{a}}\right) \\ $$$${s}^{\mathrm{6}} +\left(\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{d}}{{a}}\right){s}^{\mathrm{3}} −\frac{{b}^{\mathrm{6}} }{\mathrm{729}{a}^{\mathrm{6}} }=\mathrm{0} \\ $$$${s}=\sqrt[{\mathrm{3}}]{\frac{\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{d}}{{a}}\pm\sqrt{\left(\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{2}} }−\frac{{d}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{4}{b}^{\mathrm{6}} }{\mathrm{729}{a}^{\mathrm{6}} }}}{\mathrm{2}}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\frac{\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{d}}{{a}}\pm\sqrt{\left(\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }−\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{2}} }−\frac{{d}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{4}{b}^{\mathrm{6}} }{\mathrm{729}{a}^{\mathrm{6}} }}}{\mathrm{2}}} \\ $$$${x}={s}−{t}+\frac{{b}}{\mathrm{3}{a}} \\ $$$$ \\ $$$$ \\ $$
Commented by MJS_new last updated on 19/Nov/20
$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{Solution}\:\mathrm{is}\:{not}\:\mathrm{valid}\:\mathrm{in}\:\mathrm{these}\:\mathrm{cases} \\ $$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{0} \\ $$$$\mathrm{here}\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{Trigonometric}\:\mathrm{Solution} \\ $$$$\mathrm{as}\:\mathrm{MrW}\:\mathrm{showed} \\ $$
Commented by Tawa11 last updated on 06/Nov/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 19/Nov/20
$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{3}}{{x}}−\mathrm{1}=\mathrm{0} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} +\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} <\mathrm{0}\:\Rightarrow{three}\:{real}\:{roots} \\ $$$$\frac{\mathrm{1}}{{x}}=\mathrm{2}\:\mathrm{sin}\:\left(−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right),{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)}=\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{110}°}}\\{\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{230}°}}\\{\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{350}°}}\end{cases} \\ $$