Factor question
- To: mathgroup at smc.vnet.net
- Subject: [mg71270] Factor question
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Sun, 12 Nov 2006 06:48:22 -0500 (EST)
Why Factor gives the follwing output below?
sols = x /. Solve[x^6 - 1 == 0]
{-1, 1, -(-1)^(1/3), (-1)^(1/3), -(-1)^(2/3), (-1)^(2/3)}
Factor[x^6 - 1, Extension -> %]
(-1 + (-1)^(1/3) - x)*((-1)^(1/3) - x)*(-1 + x)*(1 + x)*(-1 +
(-1)^(1/3) + x)*((-1)^(1/3) + x)
that is why it writes e.g. (-1 + (-1)^(1/3) instead of (-1)^(2/3)
Block[{Message}, FullSimplify[(-1)^(2/3) == -1 + (-1)^(1/3)]]
True
How is possible to get with Factor the output as follows?
Times @@ Apply[Plus, Solve[x^6 - 1 == 0], 2]
(-1 + x)*(1 + x)*(-(-1)^(1/3) + x)*((-1)^(1/3) + x)*(-(-1)^(2/3) +
x)*((-1)^(2/3) + x)
or
Times @@ Apply[Plus, Flatten[Solve[x^6 - 1 == 0], 1], {1}]
(-1 + x)*(1 + x)*(-(-1)^(1/3) + x)*((-1)^(1/3) + x)*(-(-1)^(2/3) +
x)*((-1)^(2/3) + x)]
Regards
Dimitris