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I'm trying to find the roots of P(x) :$$x^3-3x\enspace-4=0$$

First I tried $P(\pm1),\enspace P(\pm2), \enspace P(\pm4)$ , and I found no rational solutions.

So I used the formula for the solutions of a third degree equation:

given $$x^3+px+q=0$$

you have $$x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}\enspace+\enspace \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$

with $p = -3,\enspace q = -4\enspace$ I found the real solution $$x = \sqrt[3]{2+\sqrt{3}}\enspace +\enspace\sqrt[3]{2 - \sqrt{3}}$$

Now how do I find the two complex solutions?

Should I divide $\enspace x^3-3x\enspace-4\enspace$ by $\Bigl(x - (\sqrt[3]{2+\sqrt{3}}\enspace +\enspace\sqrt[3]{2 - \sqrt{3}})\Bigr) $?

José Carlos Santos
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IDK
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1 Answers1

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Let $\omega=\cos\left(\frac{2\pi}3\right)+\sin\left(\frac{2\pi}3\right)i=-\frac12+\frac{\sqrt3}2i$. Then the roots of the equation are$$\sqrt[3]{2+\sqrt3}+\sqrt[3]{2-\sqrt3},\ \sqrt[3]{2+\sqrt3}\,\omega+\sqrt[3]{2-\sqrt3}\,\omega^2\text{ and }\sqrt[3]{2+\sqrt3}\,\omega^2+\sqrt[3]{2-\sqrt3}\,\omega.$$You will find an explanation here.

José Carlos Santos
  • 440,053
  • Thanks for the answer, I need some help understanding it... So I see why you choose that angle, why $|w| = 1$ and essentially multiplying the real solution by $w$ you are rotating it to get the vertices of the regular polygon which are also the complex solutions, but I miss the rule, how you did it? – IDK Aug 11 '18 at 22:47
  • I didn't choose an angle. All I did was to apply a formula (did you click on the link at the end of my answer?). The numbers $u=\sqrt[3]{2+\sqrt3}$ and $v=\sqrt[3]{2-\sqrt3}$ are such that $3uv=-p$ and that $u^3+v^3=-q$. Therefore, $u+v$ is a root. But $\omega^3=1$ and therefore $3(u\omega)(v\omega^2)=3uv=-p$ and $(u\omega)^3+(v\omega^2)^3=u^3+v^3=-q$; so, $u,\omega+v,\omega^2$ is also a root. And the same argument applies to $u,\omega^2+v,\omega$. – José Carlos Santos Aug 11 '18 at 22:55