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How to solve the equation $x^6-2\varphi^5x^5+2\varphi x+\varphi^6=0$ in radicals using Mathematica?

where $\varphi$ is the golden ratio.

Solution is Ramanujan's Class Invariant $G_{125}$, Class Invariant is always algebraic, therefore this equation is solvable in radicals.

I try to use "Solve and Reduce ", but there is no "Radical" output 

  t = (1 + Sqrt[5])/2  

  Solve[x^6 - 2 t^5 x^5 + 2 t x + t^6 == 0, x, Reals]

out

  {{x -> 1}, {x -> Root[1 - 20 #1 - 45 #1^2 - 70 #1^3 - 95 #1^4 - 118 #1^5 -   95 #1^6 - 70 #1^7 - 45 #1^8 - 20 #1^9 + #1^10 &, 2]}}
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  • $\begingroup$ @Dr.belisarius Solution is Ramanujan's Class Invariant $G_{125}$. Class Invariant is always algebraic $\endgroup$ – vito Jan 7 '16 at 16:40
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    $\begingroup$ But not every algebraic number has a radical representation. $\endgroup$ – John Doty Jan 7 '16 at 17:19
  • $\begingroup$ @JohnDoty are you sure? could you give me an example? $\endgroup$ – vito Jan 7 '16 at 17:27
  • $\begingroup$ See the "polynomial roots" bullet at en.wikipedia.org/wiki/Algebraic_number $\endgroup$ – John Doty Jan 7 '16 at 17:33
  • $\begingroup$ @JohnDoty aaah of cource.. Galois theory.. :)) $\endgroup$ – vito Jan 7 '16 at 17:45
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Only a partial answer here, but I believe your Root object is indeed solvable in radicals. I have no idea how to find the solution, but here's how I came to my conclusion.

First let's define the polynomial we're working with:

p[x_] := 1 - 20 x - 45 x^2 - 70 x^3 - 95 x^4 - 118 x^5 - 95 x^6 - 70 x^7 - 45 x^8 - 20 x^9 + x^10

Get its roots:

roots = x /. Solve[p[x] == 0, x];

Now we find a generator for the root's splitting field:

θ = ToNumberField[roots, All][[1, 1]];

Find the degree of the splitting field extension:

Exponent[θ[[1]][x], x]
(* 20 *)

From this we can find the order of the Galois group: $$ \left|\text{Gal}(\mathbb{Q}(\theta)/\mathbb{Q})\right| = [\mathbb{Q}(\theta) : \mathbb{Q}] = 20. $$

Now there's only 5 groups of order 20 and they're all solvable, so we can stop here, i.e. we can conclude $p(x)$ can be solved in radicals.

p.s. Here's info on all groups of order 20. It mentions they are all solvable.

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  • $\begingroup$ Maybe cross-list this question with math.stackexchange.com? $\endgroup$ – QuantumDot Jan 7 '16 at 23:03
  • $\begingroup$ @QuantumDot how do I cross list a question? $\endgroup$ – Chip Hurst Jan 8 '16 at 3:29
  • $\begingroup$ I have no idea. I thought you would know... $\endgroup$ – QuantumDot Jan 8 '16 at 14:03
  • $\begingroup$ @ChipHurst I think something is wrong here. As can be seen either from above or Factor[x^6 - 2 t^5 x^5 + 2 t x + t^6 /. t -> GoldenRatio, Extension -> GoldenRatio], one factor is of degree 5. If this splits then that split involves algebraic conjugates, which means it splits into subfactors all of the same degree over Q. Or so I believe. With 5 being prime this of course means there can be no such split. (But maybe I'm seeing this all wrong, what with the fog and all.) $\endgroup$ – Daniel Lichtblau Jan 8 '16 at 16:27
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    $\begingroup$ By the way I like that method of figuring the extension degree (so I upvoted). And wipe the L off your forehead, it makes you look gloomy. $\endgroup$ – Daniel Lichtblau Jan 8 '16 at 18:15

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