Let p:=(1/2 - Sqrt[23/3]/6)^(1/3) + (1/2 + Sqrt[23/3]/6)^(1/3)] = "the plastic constant", solving a tiny cubic: In[113]:= RootReduce@p Out[113]= Root[-1 - #1 + #1^3 &, 1] http://mathworld.wolfram.com/DedekindEtaFunction.html gives p = i^(-1/12) η(½+i√23/2)/√2/η(1+i√23), suggesting that η(i√23) contains (some power of) p as a factor. I've just now simplified it to DedekindEta[I Sqrt[23]] == (23^(19/24) Sqrt[(3/23)!] (( 13 (2/23)! (4/23)! (6/23)! (8/23)!)/((1/46)! (9/46)! (13/46)!))^( 1/6) ((1/23)! (9/23)!)^(1/3))/(8 2^(1/23) 3^( 2/3) ((1/2 - Sqrt[23/3]/6)^(1/3) + (1/2 + Sqrt[23/3]/6)^(1/3))^( 4/3) \[Pi]^(3/4) ((3/46)!)^(1/3)) with numerous equivalent forms, given the identity (1476395008 2^(21/23) \[Pi]^(3/2) (1/46)! ((3/46)!)^2 (5/46)! ((7/46)!)^4 (9/46)! ((11/46)!)^2 ( 13/46)! ((15/46)!)^3 (17/46)! ((19/46)!)^2 (21/46)!)/( 4853477135 Sqrt[23] ((1/23)!)^2 (2/23)! ((3/23)!)^3 (4/23)! ((5/23)!)^2 (6/ 23)! ((7/23)!)^5 (8/23)! ((9/23)!)^2 (10/23)! ((11/23)!)^3) == 1 Eric, hang on. Those η(½+i√23/2) and η(1+i√23) I sent were undersimplified! —rwg