Finding roots of sextic in radicals



Have you tried solve? This polynomial is the functional composition of
lower-degree polys so should easily be handled by solve. No need for
numerical solutions and identify. Take a look at the doc for
polydecomp if you're interested.

       -s



On 1/25/08, andre maute <andre.maute at gmx.de> wrote:
> On Thursday 24 January 2008, Andrej Vodopivec wrote:
> > On Jan 22, 2008 5:04 AM, Jordi Guti?rrez Hermoso <jordigh at gmail.com>
> wrote:
> > > If my calculations are correct, the roots of x^6 + 3*x^5 + 6*x^4 +
> > > 3*x^3 + 9*x +9 should all be expressible by radicals. In fact, they
> > > are all polynomials in terms of 2^(1/3) and a cube root of unity.
> > >
> > > How can I make Maxima tell me what the roots actually are?
> >
> > I don't know how to compute the roots in radicals, but I have some
> > code which can sometimes guess how they look like from their numerical
> > values:
> >
> > (%i1) load(identify)$
> > (%i2) x^6 + 3*x^5 + 6*x^4 + 3*x^3 + 9*x + 9$
> > (%i3) allroots(%)$
> > (%i4) identify(%);
> > (%o4)
> >
> [x=(sqrt(3)*%i)/2+(3*2^(1/3)-3/2)/3,x=(3*2^(1/3)-3/2)/3-(sqrt(3)*%i)/2,x=.2
> >250982321872863*%i+(-3/4^(1/3)-3/2)/3,x=(-3/4^(1/3)-3/2)/3-
> >
> .2250982321872863*%i,x=1.95714903975616*%i+(-3/4^(1/3)-3/2)/3,x=(-3/4^(1/3)
> >-3/2)/3-1.95714903975616*%i] (%i5) subst(first(%), %o2), ratsimp; <-- you
> > need to check if we
> > guessed correctly
> > (%o5) 0
> >
> > After you have two roots written with radicals you should be able to
> > solve the polynomial of degree 4 to get others.
> >
> > identify works mostly like the identify from maple but I think is less
> > powerfull. If you would find it interesting let me know.
>
> Perhaps you could check
> what your identify function gives for the sixth legendre polynomial.
>
> 231*x^6/16-315*x^4/16+105*x^2/16-5/16
>
> I have not seen the roots written in radicals, yet.
>
> Andre
>
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