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For the cubic equation $x^3-3x-1=0$, you can check it has three real roots by calculus knowledge. Let them be $x_1<x_2<x_3$. Now prove that $$x_3^2-x_2^2=x_3-x_1.$$ It is not difficult to check that: $$x_1=2\cos 140 ^{\circ},x_2=2\cos 100 ^{\circ},x_3=2\cos 20 ^{\circ}.$$ (Here use the formula: $\cos 3x=4\cos^3x-3\cos x.$ )

And then we can check: $$x_3^2-x_2^2=x_3-x_1.$$ I just want to know how to get: $x_1=2\cos 140 ^{\circ},x_2=2\cos 100 ^{\circ},x_3=2\cos 20 ^{\circ}.$ Is there some trick. Any help and hint will welcome.

Riemann
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  • For cubic equations, you can use Cardano's Formula: https://www.encyclopediaofmath.org/index.php/Cardano_formula – For the love of maths Oct 12 '18 at 14:06
  • See https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 and https://math.stackexchange.com/questions/2203364/solve-the-following-equation-x3-3x-sqrtx2 – lab bhattacharjee Oct 12 '18 at 14:33

2 Answers2

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Using the formula $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$ is actually quite useful, but first we have to do $x=2y$, then we get $$8y^3 - 6y - 1 = 0$$ Now we have the 4:-3 ratio and we can substitute $y=\cos(\theta)$ to get, from the above formula, $$2\cos(3\theta) = 1$$ and from there we can get cosine form of the roots.

Community
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Isaac Browne
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Since the roots are of the form $\cos(\theta)$, make the substitution $x=t+t^{-1}$ and expand to get

$$P(t+t^{-1})=\frac {t^6-t^3+1}{t^3}=\frac {t^9+1}{t^3(t^3+1)}$$ Set the above expression equal to zero and it's easy to see that$$t^9=-1$$Let $t=\cos\theta+i\sin\theta$ and expand to see that$$\theta=\frac {\pi(2k+1)}{9}$$where $k=0,2,3,\ldots$. Hence$$x=t+t^{-1}=2\cos\left(\frac {\pi(2k+1)}9\right)$$

Frank W
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