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