One I may have sent earlier, but here rewritten entirely as nth roots: (c106)block([radexpand:false],sqrt(2160^(1/6)+4)= (-sqrt(10)+2*10^(1/6)+sqrt(6)+((5^(1/3)*(sqrt(5)+sqrt(3)))/(2^(1/6))))/3) 1/6 (d106) sqrt(2160 + 4) = 1/3 1/6 5 (sqrt(5) + sqrt(3)) - sqrt(10) + 2 10 + sqrt(6) + ------------------------ 1/6 2 --------------------------------------------------------- 3 (c107) dfloat(d106) (d107) 2.75596793350447d0 = 2.75596793350447d0 Still no sign of binomial^(1/n) = six or more real terms (n>0). My 30-yr old version of Robert Maas's implementation of Rich's "number recognizer" incompletely reduces large lattices in a manner suggestive of a short-by-1 coding bug. A multiday search with Mma's LatticeReduce has turned up numerous small counterexamples to my denester, at least some of which worked subsequent to its Nov04 latest edit, so I must have bent an autoload someplace. One such poser: 1/3 1/3 5/6 5/6 1/6 1/6 sqrt(39 2 - 34 3 ) = 2 3 - 3 2 sqrt(3) + sqrt(2) 3 whose sides seem to have disproportionate coefficients. --rwg MUSTARD SPINACH PHANTASMIC SURD