       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

```

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