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
cos(2pi/11),which is a cosine value of central angle of hendecagon.(a regular 11-gon)
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
64×cos360∘(36011)6−80×cos360∘(36011)4+8×cos360∘(36011)3+24×cos360∘(36011)2−4×cos360∘(36011)−1=−0.0000000000044649
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.
Why do you think it is exact?