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Question Number 200460 by deleteduser12 last updated on 19/Nov/23

$$\mathrm{Find}\mathrm{the}{\mathrm{cardano}}^{\prime }\mathrm{s}\mathrm{solution}\mathrm{of}\mathrm{the}$$$$\mathrm{equation}{28\mathrm{x}}^3-{9\mathrm{x}}^2+1=0$$

Answered by Frix last updated on 19/Nov/23

$$28x^3-9x^2+1=0$$$$x^3-\frac{9x^2}{28}+\frac{1}{28}=0$$$$\mathrm{Before}\mathrm{using}\mathrm{any}\mathrm{other}\mathrm{method}\mathrm{we}\mathrm{try}\mathrm{all}$$$$\mathrm{factors}\mathrm{of}\pm \frac{1}{28}\Rightarrow x_1=-\frac{1}{4}.\mathrm{We}\mathrm{now}\mathrm{have}$$$$(x+\frac{1}{4})(x^2-\frac{4x}{7}+\frac{1}{7})=0$$$$\Rightarrow x_{2,3}=\frac{2}{7}\pm \frac{\sqrt{3}}{7}\mathrm{i}$$$$\mathrm{Cardano}\mathrm{means}\mathrm{more}\mathrm{work}:$$$$28x^3-9x^2+1=0$$$$x^3-\frac{9x^2}{28}+\frac{1}{28}=0$$$$x=t+\frac{3}{28}$$$$t^3-\frac{27t}{784}+\frac{365}{10976}=0\Rightarrow p=-\frac{27}{784}\wedge q=\frac{365}{10976}$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}\wedge v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$$$t_1=u+v$$$$t_2=\omega u+{\omega }^{2}v$$$$t_3={\omega }^{2}u+\omega v$$$$...$$

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