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

• To: mathgroup at smc.vnet.net
• Subject: [mg54655] Re: finding roots of 1 + 6*x - 8*x^3
• From: Scott Hemphill <hemphill at hemphills.net>
• Date: Fri, 25 Feb 2005 01:19:01 -0500 (EST)
• References: <cvk4nr\$dnj\$1@smc.vnet.net>
• Reply-to: hemphill at alumni.caltech.edu
• Sender: owner-wri-mathgroup at wolfram.com

```"Kennedy" <kennedy at oldnews.org> writes:

> 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))
>
> This monstrosity is chock full of imaginaries,
> even though I know all three roots are real.
>
> I tried Solve too but got the same thing, except
> in the form of a set of replacements. My guess is
> that Solve just calls Roots when handed a poly-
> nomial.
>
> When I ran the above through FullSimplify, I
> got three "Root" objects, the upshot of which is
> that the roots of the polynomial are indeed the
> Roots of said polynomial. Huh.
>
> What command can I use to get the roots into
> a form that are
> (a)  purely real, and
> (b)  in radical form?

What you're asking for isn't possible.  This is a topic for a number theory
course.

http://mathforum.org/library/drmath/view/53890.html

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

1 + I Sqrt[3] 1/3     1 + I Sqrt[3] 2/3
-((1 - I Sqrt[3]) (-------------)   )   (-------------)
2                     2
Out[1]= {------------------------------------- - ------------------,
4                             2

1 + I Sqrt[3] 1/3
(-------------)
2                        1
>    ------------------ + -----------------------,
2             2/3                1/3
2    (1 + I Sqrt[3])

4/3
-(1 - I Sqrt[3])        (1 + I Sqrt[3])
>    ------------------------- - ------------------}
2/3                1/3            1/3
2 2    (1 + I Sqrt[3])            4 2

In[2]:= ComplexExpand[%]

Pi                Pi                 Pi                Pi
-Cos[--]   Sqrt[3] Sin[--]           -Cos[--]   Sqrt[3] Sin[--]
9                 9        Pi        9                 9
Out[2]= {-------- - ---------------, Cos[--], -------- + ---------------}
2              2             9       2              2

Is this where you started, trying to find the cosine of Pi/9?

Scott
--
Scott Hemphill	hemphill at alumni.caltech.edu
"This isn't flying.  This is falling, with style."  -- Buzz Lightyear

```

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