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How do I write explicitly the roots of $x^3 - x+ \eta =0$? I tried online calculators but could not get any idea, also doing $x(x-1)(x+1) = -\eta$ seems is not helping too.

I tried to cleverly see for any one root since I after getting one root I can do the quadratic solving for other two roots but getting that one root is difficult as I see

lioness99a
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BAYMAX
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  • See http://www.wolframalpha.com/input/?i=x%5E3-x%2Ba%3D0 – lhf Sep 28 '18 at 10:29
  • Did you look at the Wikipedia page about the solution(s) of cubic equations ? In your case, the number of real roots will depend on $\eta$. – Claude Leibovici Sep 28 '18 at 10:30
  • There is not much way around the fact that you have to use Cardano's formula. Moreover, the derivative has two distinct real zeros, so you are guaranteed that for some values of $\eta$ you'll have $3$ (sometimes $2+1$) real roots and for some other values you'll have exactly one real root. –  Sep 28 '18 at 10:30
  • See here https://trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm – Dr. Sonnhard Graubner Sep 28 '18 at 10:30

1 Answers1

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Let

$$x=\frac2{\sqrt3}\cos\theta.$$

Then

$$\frac8{3\sqrt3}\cos^3\theta-\frac2{\sqrt3}\cos\theta+\eta=0,$$

$$4\cos^3\theta-3\cos\theta+\frac{3\sqrt3}2\eta=0,$$

so

$$\cos3\theta=-\frac{3\sqrt3}2\eta$$ gives you 3 solutions.

When

$$\left|\frac{3\sqrt3}2\eta\right|>1,$$ just use

$$x=-\text{sgn }\eta\,\frac2{\sqrt3}\cosh\theta,\\\cosh3\theta=\frac{3\sqrt3}2|\eta|.$$