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Question Number 88899 by Ar Brandon last updated on 13/Apr/20

$$Simplify$$$$\frac{\sqrt[3]{35+18i\sqrt{3}}+\sqrt[3]{35-18i\sqrt{3}}-4}{3}$$

Commented by naka3546 last updated on 14/Apr/20

$$Isituniquesolution?$$

Commented by MJS last updated on 14/Apr/20

$$\mathrm{usually}\mathrm{you}\mathrm{get}\mathrm{something}\mathrm{like}\mathrm{this}\mathrm{when}$$$$\mathrm{solving}\mathrm{a}\mathrm{polynome}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{degree}\mathrm{using}$$$$\mathrm{Cardano}\mathrm{although}D=\frac{p^3}{27}+\frac{q^2}{4}<0.\mathrm{In}\mathrm{these}$$$$\mathrm{cases}\mathrm{Cardano}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{work}$$$$-\frac{q}{2}\pm \sqrt{\frac{p^3}{27}+\frac{q^2}{4}}=35\pm \sqrt{-972}$$$$\Rightarrow p=-39\wedge q=-70$$$$x^3-39x-70=0$$$$\mathrm{trying}\mathrm{factors}\mathrm{of}70\Rightarrow$$$$\Rightarrow x_1=-5\wedge x_2=-2\wedge x_3=7$$$$\mathrm{knowing}\mathrm{that}\mathrm{the}\mathrm{angle}\mathrm{of}35+18\sqrt{3}\mathrm{i}\mathrm{is}\mathrm{within}$$$$[0;\frac{\pi }{2}]\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{must}\mathrm{be}>0$$$$\Rightarrow \sqrt[3]{35+18\sqrt{3}\mathrm{i}}+\sqrt[3]{35-18\sqrt{3}\mathrm{i}}=7$$$$\Rightarrow \mathrm{answer}\mathrm{is}1$$

Commented by Ar Brandon last updated on 14/Apr/20

Commented by Ar Brandon last updated on 14/Apr/20

$$Ithinkitgoesinallcases.Justneedto$$$$manipulatethecomplexnumbers.$$

Commented by MJS last updated on 14/Apr/20

$$\mathrm{but}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{recommended}$$$$x^3+{ax}^2+bx+c=0$$$$\mathrm{first},\mathrm{try}\mathrm{factors}\mathrm{of}c$$$$\mathrm{second},\mathrm{let}x=z-\frac{a}{3}$$$$\Rightarrow x^3+px+q=0$$$$D=\frac{p^3}{27}+\frac{q^2}{4}$$$$\Rightarrow \{\begin{array}{l}\mathrm{Cardano};D>0\\ \mathrm{trigonometric}\mathrm{solution};D<0\end{array}$$$$\mathrm{try}\mathrm{this}\mathrm{one}\mathrm{using}\mathrm{Cardano},\mathrm{exact}\mathrm{solutions}$$$$\mathrm{required}:$$$$x^3-8x^2-154x+332=0$$

Commented by Ar Brandon last updated on 02/May/20

$$\mathrm{Hi}!\mathrm{Sir}\mathrm{I}\mathrm{tried}\mathrm{and}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{getting}\mathrm{it}\mathrm{right}\mathrm{unfortunately}.$$$${\mathrm{What}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{secret}?$$

Commented by Ar Brandon last updated on 02/May/20

What PDF document would you recommend me to study in other to master the subject ? ����

Commented by MJS last updated on 02/May/20

$$\mathrm{I}\mathrm{will}\mathrm{after}\mathrm{the}\mathrm{weekend}...\mathrm{if}\mathrm{I}\mathrm{forget},\mathrm{please}$$$$\mathrm{remind}\mathrm{me}$$

Commented by Ar Brandon last updated on 02/May/20

$$\mathrm{OK}\mathrm{thanks}$$

Commented by Ar Brandon last updated on 05/May/20

Hi Sir ; just to remind you as you said. ��

Commented by MJS last updated on 05/May/20

$$\mathrm{I}\mathrm{only}\mathrm{found}\mathrm{one}\mathrm{pdf}\mathrm{in}\mathrm{German}$$$$\mathrm{but}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{article}\mathrm{on}\mathrm{Wikipedia},\mathrm{search}$$$$\mathrm{for}‘‘\mathrm{Wikipedia}\mathrm{cubic}{\mathrm{equation}}^{″}$$

Commented by Ar Brandon last updated on 05/May/20

OK, there's no problem. What German PDF did you find ? ��

Commented by MJS last updated on 05/May/20

http://mathemator.org/Media/Themen/Kubische%20Gleichungen.pdf

Commented by Ar Brandon last updated on 05/May/20

thank you Sir ��

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