lost a Mathematica 7.0 that had gone 2.5 days on
Factor[x^3-3^(3/5)+2^(1/5), Extension -> {5^(1/3),2^(1/5),3^(1/5)}]
its linear factor should be x - (3^(1/5) + 4^(1/5) - 54^(1/5) + 72^(1/5))/5^(2/3) and of course the quadratic cofactor is irreducible. mike
lost a Mathematica 7.0 that had gone 2.5 days on
Factor[x^3-3^(3/5)+2^(1/5), Extension -> {5^(1/3),2^(1/5),3^(1/5)}]
MReid> its linear factor should be
x - (3^(1/5) + 4^(1/5) - 54^(1/5) + 72^(1/5))/5^(2/3)
and of course the quadratic cofactor is irreducible.
mike
Thereby reenacting one of my favorite denestings: 1/5 3/5 3/5 2/5 1/5 2/5 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 (3 - 2 ) = ------------------------------------- , 2/3 5 a major simplification that looks anything but. For greater intrigue, I have numerous cases of sqrt(binomial) = pentanomial, but *never* anything higher, except complex. Are there any real ones?? --rwg NTH ROOT SHORT TON LOCUST BEAN UNCOUNTABLES
On 12/17/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
... Thereby reenacting one of my favorite denestings:
1/5 3/5 3/5 2/5 1/5 2/5 3/5 1/5 1/3 - 2 3 + 2 3 + 3 + 2 (3 - 2 ) = ------------------------------------- , 2/3 5
a major simplification that looks anything but.
Indeed! u := 3^(1/5); v := 2^(1/5); w := 5^(1/3); l := (u^3 - v)^1/3; r := (u + v^2 - v*u^3 + v^3*u^2)/w^2; evalf(l - r); # -.6607822845 ?! WFL
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Fred lunnon -
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rwg＠sdf.lonestar.org