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Question Number 175095 by Gbenga last updated on 18/Aug/22

$${\mathbf{x}}^3+{\mathbf{x}}^2+\mathbf{x}+\mathbf{c}=0$$$$\mathbf{find}\mathbf{x}$$

Answered by MJS_new last updated on 18/Aug/22

$$x=z-\frac{1}{3}$$$$z^3+\frac{2}{3}z+(c-\frac{7}{27})=0$$$$\Rightarrow$$$$z_1=u+v$$$$z_2=\omega u+{\omega }^{2}v$$$$z_3={\omega }^{2}u+\omega v$$$$\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}\wedge v=-\sqrt[3]{\frac{q}{2}\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$$$p=\frac{2}{3}\wedge q=c-\frac{7}{27}$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=\frac{c^2}{4}-\frac{7c}{54}+\frac{1}{36}>0\mathrm{\forall }c\in \mathbb{R}$$$$\mathrm{do}\mathrm{the}\mathrm{rest}\mathrm{yourself}$$

Commented by Gbenga last updated on 19/Aug/22

Commented by Tawa11 last updated on 19/Aug/22

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

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