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 >