       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)/2)^(1/3)/2 +
>    1/(2^(2/3)*(1 + I*Sqrt)^(1/3)) ||
>  x == -((1 - I*Sqrt)*((1 + I*Sqrt)/2)^(1/3))/4 -
>    ((1 + I*Sqrt)/2)^(2/3)/2 ||
>  x == -(1 - I*Sqrt)/(2*2^(2/3)*(1 + I*Sqrt)^
>       (1/3)) - (1 + I*Sqrt)^(4/3)/(4*2^(1/3))

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

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

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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