9

3

This question was originally asked by @fsrong70 six months ago. The OP deleted it shortly after posting and has not returned to this site since. I had just figured it out when it was deleted. I waited to see if the OP would repost it, but not yet. So I'm posting it with my solution.

Considering the solution to the following equation

$x^7+x^6-18 x^5-35 x^4+38 x^3+104 x^2+7 x-49=0$

```
poly = x^7 + x^6 - 18 x^5 - 35 x^4 + 38 x^3 + 104 x^2 + 7 x - 49;
```

This problem could be solved using notions from Galois theory and the Galois group of a polynomial, and the 43rd root of unity ($\displaystyle e^\frac{2i\pi}{43}$) .

The solutions are expressed in terms of trigonometric (cosine) functions. The seven solutions are:

$\displaystyle \alpha_1 = 2\cos(\frac{2\pi}{43}) + 2\cos(\frac{12\pi}{43}) + 2\cos(\frac{14\pi}{43})$

$\displaystyle \alpha_2 =2\cos(\frac{4\pi}{43}) + 2\cos(\frac{24\pi}{43}) + 2\cos(\frac{28\pi}{43})$

$\displaystyle \alpha_3 = 2\cos(\frac{6\pi}{43}) + 2\cos(\frac{36\pi}{43}) + 2\cos(\frac{42\pi}{43})$

$\displaystyle \alpha_4 =2\cos(\frac{8\pi}{43}) + 2\cos(\frac{30\pi}{43}) + 2\cos(\frac{38\pi}{43})$

$\displaystyle \alpha_5 =2\cos(\frac{10\pi}{43}) + 2\cos(\frac{16\pi}{43}) + 2\cos(\frac{24\pi}{43})$

$\displaystyle \alpha_6 =2\cos(\frac{18\pi}{43}) + 2\cos(\frac{22\pi}{43}) + 2\cos(\frac{40\pi}{43})$

$\displaystyle \alpha_7 =2\cos(\frac{20\pi}{43}) + 2\cos(\frac{32\pi}{43}) + 2\cos(\frac{34\pi}{43})$

Might this problem be entirely solved and the symbolic solutions computed with Mathematica?

This problem was asked and solved on Quora, but not in relation to *Mathematica*.

2Note that substituting the roots into

`poly`

and using`Simplify`

to see if they satisfy it takes a very long time. (Using`N`

after substituting gives verification almost immediately, rhough.) – murray – 2019-01-12T20:59:51.750@murray Yes, it does. This is a bit faster:

`FullSimplify @ TrigToExp@Activate[poly /. roots]`

. – Michael E2 – 2019-01-12T21:01:05.623