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Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$

I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.

I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, and using Vieta's trigonometric solution formula, I got $$x_1=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {\arccos \frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_2=\frac {2\sqrt{31}}3\cdot\cos\left(\frac {2\pi+\arccos\frac {2}{\sqrt{31}}}3\right)-\frac 13\\x_3=\frac {2\sqrt{31}}{3}\cdot\cos\left(\frac {4\pi+\arccos\frac 2{\sqrt{31}}}3\right)-\frac 13$$ But that's not the form I want. I'm looking for a form of $2\left(\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}+\cos\frac {\text{something}}{31}\right)$.

I've spent so much time, that I'm practically burnt out. -.-

Frank
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2 Answers2

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If you wish to express as roots of cubics the sums of trigonometric functions with arguments that use a prime of form $p=6\color{blue}n+1$, then the number of addends is $\color{blue}n$. Thus, the reason you couldn't find $p=31$ was that you should have used $\color{blue}5$ addends.

$p=31$

A root of $$x^3+x^2-(2\times\color{blue}5)x-8=0\tag1$$ (note the $5$) is given by $$x_1 = 2\left(\cos\tfrac {2\pi}{31}+\cos\tfrac {4\pi}{31}+\cos\tfrac {8\pi}{31}+\cos\tfrac {16\pi}{31}+\cos\tfrac {32\pi}{31}\right) =3.083872\dots$$ More succinctly, let $\displaystyle\beta=\frac{2\pi}{31}$, then, $$x_1=2\sum_{k=1}^5\cos\big(2^k\times\beta\big)=3.083872\dots\\x_2=2\sum_{k=1}^5\cos\big(2^k\times3\beta\big)=-0.786802\dots\\x_3=2\sum_{k=1}^5\cos\big(2^k\times5\beta\big)=-3.29707\dots$$

$p=43$

Similarly, given the cubic, $$x^3+x^2-(2\times\color{blue}7)x+8=0\tag2$$ Let $\displaystyle\gamma=\frac{2\pi}{43}$, then the roots are, $$x_1=2\sum_{k=1}^7\cos\big(2^k\times\gamma\big)=2.88824\dots\\x_2=2\sum_{k=1}^7\cos\big(2^k\times3\gamma\big)=0.615072\dots\\x_3=2\sum_{k=1}^7\cos\big(2^k\times7\gamma\big)=-4.50331\dots$$ though not all primes $p=6n+1$ will have such neat cubic roots. The family $p=31,43,109,\dots$ is discussed in this post.

P.S. See also mercio's general answer here which uses cosets.

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Tito Piezas III
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  • I'm wondering, but what would you do if it was a polynomial whose solutions could not be expressed with the $\text{p}^{\text{th}}$ root of unity? Something like the famous $\sqrt{4+\sqrt{4+\sqrt{4-x}}}=x$ problem: $x^3+x^2-6x-7=0$? – Frank Oct 26 '16 at 02:42
  • Also, where do you get all these amazing theorems and such? Do you find them yourself, or lots of reading and brainstorming??? – Frank Oct 26 '16 at 02:43
  • @Frank: Actually, that polynomial can still be expressed by the $\text{p}^{\text{th}}$ root of unity though $p=6n+1=19$ is one of those infinitely many primes that don't belong to the neat family above. For example, let $\beta = 2\pi/19$ then $$x_1=2\left(\cos\big(2\beta\big) +\cos\big(3\beta\big)+\cos\big(5\beta\big)\right)$$ with the multipliers probably also dependent on cosets. – Tito Piezas III Oct 26 '16 at 03:32
  • @Frank: Reading the literature, and some I find myself with the help of Mathematica. :) – Tito Piezas III Oct 26 '16 at 03:34
  • @TitoPiezasIII Using your method, what would the other $2$ roots be then? –  Oct 26 '16 at 20:45
  • @user332252: Didn't you mention the case $p=19$ in this post? (All three roots are there.) – Tito Piezas III Oct 27 '16 at 01:14
  • @TitoPiezasIII Yes, but I'm wondering what would the forms look like in $\beta=\frac {2\pi}{19}$ form. –  Oct 27 '16 at 02:58
  • @user332252: For $p=19$, let $\beta = \color{blue}{\frac{\pi}{19}}$. Then the roots of $x^3+x^2-6x-7=0$ are $$x_1=2\big(\cos(2\times2\beta)+\cos(3\times2\beta)+\cos(5\times2\beta)\big)\x_2=2\big(\cos(2\times2^2\beta)+\cos(3\times2^2\beta)+\cos(5\times2^2\beta)\big)\x_3=2\big(\cos(2\times2^3\beta)+\cos(3\times2^3\beta)+\cos(5\times2^3\beta)\big)$$ Note that the last multiplier is the sum of the first two. (I changed the value in the other link. The periodicity of trigonometric functions allows these aesthetic changes.) – Tito Piezas III Oct 27 '16 at 03:49
  • @TitoPiezasIII The roots of $x^3+3x^2-4x-13=0$ are$$x_1=2\left(\cos\frac {4\pi}7+\cos\frac {6\pi}{7}+\cos\frac {10\pi}7\right)\ x_2=2\left(\cos\frac {8\pi}7+\cos\frac {12\pi}7+\cos\frac {20\pi}7\right)\ x_3=2\left(\cos\frac {16\pi}7+\cos\frac {24\pi}7+\cos\frac {40\pi}7\right)$$ – Frank Oct 27 '16 at 15:22
  • @Frank: Interesting. It has exactly the same form as for $p=19$. Let $\beta =\frac{\pi}{7}$, then the roots of $x^3+3x^2-4x-13=0$ are $$x_k=2\big(\cos(2\times2^k\beta)+\cos(3\times2^k\beta)+\cos(5\times2^k\beta) \big)$$ for $k=1,2,3$. – Tito Piezas III Oct 27 '16 at 15:36
  • @TitoPiezasIII Just wondering, and this may be off topic, but do you know where all the nested radicals stuff is located in Ramanujan's notebooks by Bruce C. Berndt? – Frank Oct 27 '16 at 19:42
  • @Frank: I have copies of all five books, but I forget where those nested radicals are precisely. Some are scattered about, like the beautiful $$\sqrt[8]{1+\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2},\frac{1+\sqrt[4]{5}}{\sqrt{2}}$$ which is in Vol 5, page 300. – Tito Piezas III Oct 28 '16 at 01:56
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1) Reduce the standard way your equation $x^3+x^2-10x-8=0$ to the form $$x^3+ax+b=0\qquad(*)$$

2) Since you have the identity $$4\cos^3\theta-3\cos \theta-\cos 3\theta=0\qquad(**)$$ make $x=u\cos\theta$ in $(*)$ so you get $$u^3\cos^3\theta+au\cos\theta+b=0\iff4\cos^3\theta+\frac{4au}{u^3}\cos\theta+\frac{4b}{u^3}\qquad(***)$$

3) In order to refer $(***)$ to $(**)$ you need to take $$-3=\frac{4au}{u^3}\iff u=2\sqrt{\frac{-2a}{3}}$$ and $$\frac{4b}{u^3}=-\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$ which gives $$\cos3\theta=\frac{3b}{2a}\sqrt{\frac{-3}{a}}$$

Hence your roots are given by $$x_k=2\sqrt{\frac{-a}{3}}\cos\frac{\theta_k}{3}$$ where $$\theta_k=\arccos\left(\frac{3b}{2a}\sqrt{\frac{-3}{a}}-\frac{2k\pi}{3}\right);\space k=0,1,2$$

Note.-You can use this method because the three roots of your equation are real.

Piquito
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