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MathGroup Archive 2005

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Re: finding roots of 1 + 6*x - 8*x^3

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54681] Re: finding roots of 1 + 6*x - 8*x^3
  • From: Curt Fischer <tentrillion at gmail.NOSPAM.com>
  • Date: Fri, 25 Feb 2005 01:20:35 -0500 (EST)
  • References: <cvk4nr$dnj$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Kennedy wrote:

ComplexExpand works somewhat in your case.

> Hello All,
> 
> I am trying to find the roots of
> 1 + 6*x - 8*x^3.
> 
> Roots[1+6*x-8*x^3==0,x] yields this ugly thing:
> (made uglier by my converting to InputForm)
> 
> x == ((1 + I*Sqrt[3])/2)^(1/3)/2 +
>    1/(2^(2/3)*(1 + I*Sqrt[3])^(1/3)) ||
>  x == -((1 - I*Sqrt[3])*((1 + I*Sqrt[3])/2)^(1/3))/4 -
>    ((1 + I*Sqrt[3])/2)^(2/3)/2 ||
>  x == -(1 - I*Sqrt[3])/(2*2^(2/3)*(1 + I*Sqrt[3])^
>       (1/3)) - (1 + I*Sqrt[3])^(4/3)/(4*2^(1/3))

In[86]:=
solHmmm = ComplexExpand[x /. Solve[1 + 6*x - 8*x^3 == 0, x]]

Out[86]=
{Cos[Pi/9],
(-(1/2))*Cos[Pi/9] - (1/2)*Sqrt[3]*Sin[Pi/9],
(-(1/2))*Cos[Pi/9] + (1/2)*Sqrt[3]*Sin[Pi/9]}

> What command can I use to get the roots into
> a form that are
> (a)  purely real, and

The above expression is purely real, but

> (b)  in radical form?

It's not in radical form.  How can you represent Sines and Cosines of 
Pi/9 in radical form?  I'd be interested in a method that does so.

-- 
Curt Fischer


> 
> Thanks,
> Kennedy
> 
> PS.  Mathematica 4.2
> 


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