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Question Number 61923 by naka3546 last updated on 12/Jun/19

$$Findallsolutionsof$$$$x^3-12x+8=0$$

Answered by MJS last updated on 12/Jun/19

$$x^3+px+q=0$$$$$$$$p=-12$$$$q=8$$$$D=\frac{p^3}{27}+\frac{q^2}{4}=-48<0\Rightarrow \mathrm{trigonometric}\mathrm{method}$$$$x_k=\frac{2\sqrt{-3p}}{3}\sin (\frac{2\pi k}{3}+\frac{1}{3}\arcsin \frac{3\sqrt{3}q}{2\sqrt{-p^3}})\mathrm{with}k=0,1,2$$$$x_0=4\sin \frac{\pi }{18}$$$$x_1=4\cos \frac{2\pi }{9}$$$$x_2=-4\cos \frac{\pi }{9}$$

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