# [Maxima] Finding roots of sextic in radicals

Barton Willis willisb at unk.edu
Wed Jan 23 12:24:51 CST 2008

A to_poly amusement:

(%i118) x=-(2^(1/3)*(sqrt(3)*%i+1)+sqrt(3)*%i+1)/2;
(%o118) x=(-2^(1/3)*(sqrt(3)*%i+1)-sqrt(3)*%i-1)/2

(%i119) to_poly(%);
(%o119)
[[x-(-(%i*%g33+1)*%g34-%i*%g33-1)/2,2=%g34^3,3=%g33^2],[-%pi/3<carg(%g34),carg(%g34)<=%pi/3,-%pi/2<carg(%g33),
carg(%g33)<=%pi/2]]

(%i120) elim_allbut(first(%),[x]);
(%o120)
[[x^6+3*x^5+6*x^4+3*x^3+9*x+9],[x^2+%g34*x+x+%g34^2+2*%g34+1,%g33^2-3]]

Barton

maxima-bounces at math.utexas.edu wrote on 01/23/2008 12:15:51 PM:

> On 22/01/2008, Stavros Macrakis <macrakis at alum.mit.edu> wrote:
> >     p: x^6+3*x^5+6*x^4+3*x^3+9*x+9;
> >     p2: factor(p,q^3-2)\$
> >     solve(subst(q=2^(1/3),p2),x);
>
> Sweet, I didn't know you could give hints to the factor function.
> Good to know.
>
> As for those of you who offered numerical solutions, yes, there is no
> general formula for a quintic or higher, but this wasn't a general
> polynomial, but a very specific one. It's the minimal polynomial for
> 2^(1/3) + exp(2*%pi*%i/3) over the rationals. :-)
>
> Thank you for your help,
> - Jordi G. H.
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