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Question Number 56368 by Hassen_Timol last updated on 15/Mar/19

$$Whatisthecommonformulatoobtain$$$$the3solutionsofapolynomialequation$$$$inthefollowingform?$$$$$$$${ax}^3+b^{2}x+cx+d=0$$

Commented by mr W last updated on 15/Mar/19

Commented by Hassen_Timol last updated on 16/Mar/19

$$Thankyousomuch!!!!!$$

Answered by MJS last updated on 15/Mar/19

$$1.\mathrm{divide}\mathrm{by}a$$$$x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}=0$$$$2.\mathrm{substitute}x=z-\frac{b}{3a}$$$$z^3+\frac{3ac-b^2}{3a^2}z+\frac{27a^{2}d-9abc+2b^3}{27a^3}=0$$$$3.\mathrm{substitute}\frac{3ac-b^2}{3a^2}=p;\frac{27a^{2}d-9abc+2b^3}{27a^3}=q$$$$z^3+pz+q=0$$$$4.\mathrm{check}\mathrm{resolvability}$$$$D=\frac{p^3}{27}+\frac{q^2}{4}$$$$D⩾0\Rightarrow {\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{formula}$$$$D<0\Rightarrow \mathrm{trigonometric}\mathrm{formula}$$$$5.\mathrm{solve}$$$$\mathrm{Cardano}:$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}};v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{essential}\mathrm{to}\mathrm{take}\mathrm{the}\mathrm{principal}{3}^{\mathrm{rd}}\mathrm{root}$$$$z_1=u+v$$$$z_2=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v$$$$z_3=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v$$$$\mathrm{trigonometric}:$$$$z_n=\frac{2}{3}\sqrt{-3p}\sin (\frac{2\pi }{3}n+\frac{1}{3}\arcsin \frac{3\sqrt{3}q}{2\sqrt{-p^3}})\mathrm{with}n=1,2,3$$$$$$

Commented by Hassen_Timol last updated on 16/Mar/19

$$Thanksalot...$$

Commented by mr W last updated on 16/Mar/19

$$verynicesummary!thankssir!$$

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