Rich, note that throwing in a factor of 25 reduces many days to Timing[Factor[x^3-(25*3^(3/5)-25*2^(1/5)),Extension->{2,3}^(1/5)]]

{0.297, (2^(2/5) + 3^(1/5) + 2^(3/5) 3^(2/5) - 2^(1/5) 3^(3/5) - x) (5 2^(4/5) - 5 2^(2/5) 3^(1/5) - 5 3^(2/5) + (-2^(2/5) - 3^(1/5) - 2^(3/5) 3^(2/5) + 2^(1/5) 3^(3/5)) x - x^2)}

< 1/3 sec. ! ?

that's what i did (after mistakenly futzing around with 5 * (3^(3/5) - 2^(1/5)) for a spell). with Pari/gp ( http://pari.math.u-bordeaux.fr/ ) i needed to initialize the number field "nfinit" (0.305 secs) and then the factorization took 0.224 secs. so time is comparible. i am currently running "nfinit" on the degree 75 field Q(2^(1/5), 3^(1/5), 5^(1/3)) and it has been going for over 24 hours. if it finishes, i'll try the factorization without the catalyzing factor of 25 . mike p.s. i should have realized the easy way to do it is to use the cheat sheet at http://www.tweedledum.com/rwg/identsi2.gif !