18 Dec
2008
18 Dec
'08
12:33 p.m.
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)}] [rwg:]
On 12/18/08, Phil Carmody <thefatphil@yahoo.co.uk> wrote:
.... So far, I've only seen the numeric verifications, but surely a symbolic one would be more appropriate?
? f2=Mod(a,a^5-2) Mod(a, a^5 - 2) ? f3=Mod(b,b^5-3) Mod(b, b^5 - 3)
I cubed both sides, which is fair as we're entirely within the reals.
? lift(lift((-f2*f3^3 + f2^3*f3^2 + f3 + f2^2)^3)) -25*a + 25*b^3 ? lift(lift((f3^3-f2)*25)) -25*a + 25*b^3
Well, I didn't have 2.5 days to spare --- and anyway, I couldn't find my spectacles! WFL