4 ArcCos[r/(1 + r)] + ArcCos[1 - 2/(1 + r)^2] - 4 ArcCos[(1 - (-2 + r) r)/(1 + r)^2] - 4 ArcSec[1 + r] /. r -> Root[-1 - 2 #1 + 3 #1^2 + 8 #1^3 &, 1, 0]; Sign@% N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating 4 ArcCos[Root[-1+Times[<<2>>]+Times[<<2>>]+Times[<<2>>]&,1,0]/(1+Root[<<1>>&,1,0])]-4 ArcCos[(1-(-2+Root[<<3>>]) Root[<<1>>&,1,0])/(1+Root[<<1>>&,1,0])^2]+ArcCos[1-2/(1+Root[<<1>>&,1,0])^2]-4 ArcSec[1+Root[-1+Times[<<2>>]+Times[<<2>>]+Times[<<2>>]&,1,0]]. Translation: It's zero. (Note how it helpfully abbreviates Root[-1 - 2 #1 + 3 #1^2 + 8 #1^3 &, 1] to Root[-1+Times[<<2>>]+Times[<<2>>]+Times[<<2>>]&,1,0]].) After several days, FullSimplify@%% returned -((3 \[Pi])/2) - 4 ArcSin[Root[4 + 9 #1 + 2 #1^2 + #1^3 &, 1]] + I Log[Root[1 - 2048 #1 + 1460739 #1^2 + 2633728 #1^3 + 1460739 #1^4 - 2048 #1^5 + #1^6 &, 1]] But hey, look at this! In[116]:= %72 == {1, I}.RootReduce@ReIm@%72 Out[116]= Root[1 - 2048 # + 1460739 #^2 + 2633728 #^3 + 1460739 #^4 - 2048 #^5 + #^6& , 1, 0] == Root[329728 + 365184 # - 1024 #^2 + #^3& , 1, 0] + I Root[159103 + 365827 # - 643 #^2 + #^3& , 1, 0] —rwg Kids: If R is a root of a palindromic (= "reciprocal") polynomial, then so is 1/R. Why? Uh oh, Mathematica can't Solve[x^6 + x^5 + x + 1 == 5 x^3] (in radicals). Help it out! —Bill