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Question Number 215177 by Ismoiljon_008 last updated on 30/Dec/24

$$$$$$x^3+x^2+x=-\frac{1}{3}$$$$\mathcal{F}indthevalueofx!$$$$$$$$\mathcal{H}elpme,please$$$$$$

Answered by mnjuly1970 last updated on 30/Dec/24

$$3x^3+3x^2+3x+1=0$$$$2x^3+{(x+1)}^3=0$$$$(\sqrt[3]{2}x+x+1)(\{\sqrt[3]{4}{x}^2-x^2\sqrt[3]{2}-\sqrt[3]{2}x+x^2+2x+1\}>0)=0$$$$-x(\sqrt[3]{2}+1)=1\Rightarrow x=\frac{-1}{1+\sqrt[3]{2}}$$

Answered by Ghisom last updated on 31/Dec/24

$$x^3+x^2+x+\frac{1}{3}=0$$$$x=t-\frac{1}{3}$$$$t^3+\frac{2}{3}t+\frac{2}{27}=0$$$${\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{Formula}\mathrm{gives}$$$$t_1=-\frac{2^{2/3}}{3}+\frac{2^{1/3}}{3}$$$$\Rightarrow x_1=-\frac{2^{2/3}-2^{1/3}+1}{3}$$$$t_2=-\frac{2^{2/3}}{3}\omega +\frac{2^{1/3}}{3}{\omega }^2$$$$\Rightarrow x_2=\frac{2^{2/3}-2^{1/3}-2}{6}+\frac{(2^{2/3}+2^{1/3})\sqrt{3}}{6}\mathrm{i}$$$$t_3=-\frac{2^{2/3}}{3}{\omega }^2+\frac{2^{1/3}}{3}\omega$$$$\Rightarrow x_3=\frac{2^{2/3}-2^{1/3}-2}{6}-\frac{(2^{2/3}+2^{1/3})\sqrt{3}}{6}\mathrm{i}$$

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