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Question Number 122783 by mathocean1 last updated on 19/Nov/20

$$solve:$$$$x^3+3x^2-1=0$$

Commented by Dwaipayan Shikari last updated on 19/Nov/20

$$x=y-1$$$${(y-1)}^3+3{(y-1)}^2-1=0$$$$y^3-3y^2+3y-1+3y^2-6y+3-1=0$$$$y^3-3y=-1$$$$y=s-t$$$${(s-t)}^3-3(s-t)=-1$$$$s^3-t^3-3st(s-t)-3(s-t)=-1$$$$st=-1$$$$s^3-t^3=-1\Rightarrow -\frac{1}{t^3}-t^3=-1\Rightarrow t^6-t^3+1=0\Rightarrow t^3=\frac{1\pm i\sqrt{3}}{2}$$$$t=\sqrt[3]{\frac{1\pm i\sqrt{3}}{2}}$$$$s=\sqrt[3]{\frac{-1\pm i\sqrt{3}}{2}}$$$$x=\sqrt[3]{\frac{-1\pm i\sqrt{3}}{2}}-\sqrt[3]{\frac{1\pm i\sqrt{3}}{2}}-1$$

Commented by Dwaipayan Shikari last updated on 19/Nov/20

$$Generally$$$${ax}^3+{bx}^2+cx+d=0$$$$thendepressedequation$$$$substitutex=y-\frac{b}{3a}$$$$a{(y-\frac{b}{3a})}^3+b{(y-\frac{b}{3a})}^2+c(y-\frac{b}{3a})+d=0$$$${ay}^3-by(y-\frac{b}{3a})-\frac{b^3}{27a^2}+{by}^2-\frac{2b^{2}y}{3a}+\frac{b^3}{9a^2}+cy-\frac{bc}{3a}+d=0$$$${ay}^3+\frac{b^{2}y}{3a}+\frac{2b^3}{27a^2}-\frac{2b^{2}y+bc}{3a}+d=0$$$$y^3-y\frac{b^2}{3a^2}+(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})=0$$$$y=s-t$$$$3st=\frac{b^2}{3a^2}\Rightarrow t=\frac{b^2}{9a^{2}s}$$$$s^3-t^3=-(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})$$$$s^3-\frac{b^6}{729a^6{s}^3}=(\frac{bc}{3a^2}-\frac{2b^3}{27a^3}-\frac{d}{a})$$$$s^6+(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a})s^3-\frac{b^6}{729a^6}=0$$$$s=\sqrt[3]{\frac{\frac{bc}{3a^2}-\frac{2b^3}{27a^3}-\frac{d}{a}\pm \sqrt{{(\frac{bc}{3a^2}-\frac{2b^3}{27a^2}-\frac{d}{a})}^2+\frac{4b^6}{729a^6}}}{2}}$$$$t=\sqrt[3]{\frac{\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}\pm \sqrt{{(\frac{bc}{3a^2}-\frac{2b^3}{27a^2}-\frac{d}{a})}^2+\frac{4b^6}{729a^6}}}{2}}$$$$x=s-t+\frac{b}{3a}$$$$$$$$$$

Commented by MJS_new last updated on 19/Nov/20

$${\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{Solution}\mathrm{is}not\mathrm{valid}\mathrm{in}\mathrm{these}\mathrm{cases}$$$$x^3+px+q=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}<0$$$$\mathrm{here}\mathrm{we}\mathrm{need}\mathrm{the}\mathrm{Trigonometric}\mathrm{Solution}$$$$\mathrm{as}\mathrm{MrW}\mathrm{showed}$$

Commented by Tawa11 last updated on 06/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 19/Nov/20

$$\frac{1}{x^3}-\frac{3}{x}-1=0$$$${(-1)}^3+{(-\frac{1}{2})}^2<0\Rightarrow threerealroots$$$$\frac{1}{x}=2\sin (-\frac{1}{3}{\sin }^{-1}\frac{1}{2}+\frac{2k\pi }{3})$$$$=2\sin (-\frac{\pi }{18}+\frac{2k\pi }{3}),k=0,1,2$$$$\Rightarrow x=\frac{1}{2\sin (-\frac{\pi }{18}+\frac{2k\pi }{3})}=\{\begin{array}{l}\frac{1}{2\sin 110°}\\ \frac{1}{2\sin 230°}\\ \frac{1}{2\sin 350°}\end{array}$$

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