Yay,I got it.One of the soloutions of the equation 64x^6-80x^4+8x^3+24x^2-4x-1=0 is the exact value of

This is what I get, numerically, using MathCad:

.

Well that is great :)

64x^6-80x^4+8x^3+24x^2-4x-1=0

64*(cos(2pi/11))^6-80*(cos(2pi/11))^4+8*(cos(2pi/11))^3+24*(cos(2pi/11))^2-4*(cos(2pi/11))-1

$${\mathtt{64}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{360}}}{{\mathtt{11}}}}\right)}}^{\,{\mathtt{6}}}{\mathtt{\,-\,}}{\mathtt{80}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{360}}}{{\mathtt{11}}}}\right)}}^{\,{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{360}}}{{\mathtt{11}}}}\right)}}^{\,{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{24}}{\mathtt{\,\times\,}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{360}}}{{\mathtt{11}}}}\right)}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{360}}}{{\mathtt{11}}}}\right)}{\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.000\: \!000\: \!000\: \!004\: \!464\: \!9}}$$

Yes it looks like you might be right - it is certainly is very close to zero :)

I am not so sure that Wolfram|Alpha agrees with you though.  

http://www.wolframalpha.com/input/?i=%2864x%5E6-80x%5E4%2B8x%5E3%2B24x%5E2-4x-1%29%2F%28x-cos%282*pi%2F11%29

Why do you think it is exact?

Thanks Alan,

I'm still scratching my head on this one. (contemplation and confusion - not nits   LOL )