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