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Do you know a way to solve exactly a general sixth order polynomial equation:

$x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=0$ ?

According to this link, it is possible to solve it in terms of Kampé de Fériet functions, but in the bibliography they provide I do not find any explicit reference to these functions. I have also found this method, but the fact it has zero citations makes me suspicious. I know there are solutions for special cases, but I would like to know if there is a general formula for every $a_n$. Thanks in advance.

P.S.: here the Wiki says that the method of differential resolvents "may also be generalized to equations of arbitrarily high degree", but the references are written in German :-(

user2983638
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  • the general solution is there up to fourth order – E.H.E Jun 05 '15 at 10:17
  • In what form do u want the solution - series, special functions, limit or anything else? Some computer algebra systems already have root of a polynomial as a built-in function. – Anixx Jun 05 '15 at 10:17
  • Galois showed that there is no general solution in terms of radicals to a polynomial of order greater than 4 – danimal Jun 05 '15 at 10:18
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    I think the O.P. doesn't ask specifically for a solution using only roots and rational functions. – Bernard Jun 05 '15 at 10:21
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    @Essam: the general solution has been found up to order 5, in terms of Bring radicals, hypergeometric functions or elleptic functions link and link – user2983638 Jun 05 '15 at 10:22
  • @ Anixx & danimal: I know the solutions cannot be expressed in terms of radicals. Series and special functions would be perfect! – user2983638 Jun 05 '15 at 10:25

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I have a sketch of a possible solution to the problem. In the case of a quintic polynomial, you have to use a Tschirnhaus transformation in order to transform the generic equation $x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0$ to a simpler form $x^5+x+a=0$. Then you can solve this last equation by

1) defining $f(x)=x^5+x$;

2) observing that $x=f^{-1}(-a)$;

3) expanding $f^{-1}(\cdot )$ in a Taylor series

as explained here, which gives the roots in terms of hypergeometric functions. In the case of the sextic equation, since in 1834 G. B. Jerrard showed that a Tschirnhaus transformation can be used to eliminate the $x^{n-1}$, $x^{n-2}$, and $x^{n-3}$ terms for a general polynomial equation of degree $n>3$, this means that it is possible to transform the original sextic equation to something like $x^6+x^2+ax+b=0$. If now we use the same trick explained above, I think that somehow we can invert the function, but this time we have two parameters, i.e. $a$, $b$. That's why we need something like the Kampé de Fériet functions, which indeed are two-variable generalization of the hypergeometric series. But I'm just guessing...

user2983638
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  • Yes, you are correct. The general sextic first has to be reduced to the form $x^6+x^2+ax+b = 0$, then this is the one solved by the Kampé de Fériet functions. Unfortunately, I've never seen a specific example in all this time. – Tito Piezas III Jun 23 '23 at 17:29