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Re: Re: FindRoot

  • To: mathgroup at smc.vnet.net
  • Subject: [mg98769] Re: [mg98744] Re: [mg98693] FindRoot
  • From: Gerry Flanagan <flanagan at materials-sciences.com>
  • Date: Sat, 18 Apr 2009 03:39:36 -0400 (EDT)
  • References: <200904170830.EAA23546@smc.vnet.net>

After the solution below, maybe you'll want
ComplexExpand[r] // Simplify

Gerry F.

Bob Hanlon wrote:
> eqn = x^3 - 4 a^2 x - 2 a^3 == 0;
>
> r = Simplify[x /. Solve[eqn, x], a > 0]
>
> {((4*3^(1/3) + (9 + I*Sqrt[111])^
>              (2/3))*a)/(3^(2/3)*
>         (9 + I*Sqrt[111])^(1/3)), 
>    (I*(-12 + 4*I*Sqrt[3] + 
>            I*3^(1/6)*(9 + I*Sqrt[111])^
>                (2/3) + (27 + 3*I*Sqrt[111])^
>              (2/3))*a)/(2*3^(5/6)*
>         (9 + I*Sqrt[111])^(1/3)), 
>    ((12*I - 4*Sqrt[3] - 
>            3^(1/6)*(9 + I*Sqrt[111])^
>                (2/3) - 
>            I*(27 + 3*I*Sqrt[111])^(2/3))*
>         a)/(2*3^(5/6)*(9 + I*Sqrt[111])^
>           (1/3))}
>
> Plot[r, {a, 0, 5}]
>
>
> Bob Hanlon
>
> ---- Miguel <misvrne at gmail.com> wrote: 
>
> =============
> Hi all,
> How can I to find the reals roots of a cubic equation in simbolic
> form: For example,
>
> Find the roots of
> x^3-4a^2x-2a^3==0
>
> where "a" is real and a>0.
>
> Thanks
>
>
>
>
>   



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