Re: Roots of x^3 = 1;
From: Virgil (ITSnetNOTcom#virgil_at_COMCAST.com)
Date: 10/27/04
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Date: Wed, 27 Oct 2004 00:20:54 -0600
In article <nTFfd.10706$ta5.6947@newsread3.news.atl.earthlink.net>,
Kira Yamato <no@mail.com> wrote:
> Saju wrote:
> > Hello,
> >
> > I was teaching myself some algebra when I got stuck at the following
> > ...
> >
> > I am trying to find roots of x^3 = 1
> >
> > x =1 is a solution. But keeping x = i^(4/3) also seems to work, so
> > does i^(8/3), i^(16/3) and so on - I am able to generate 1 real root
> > and infinite complex roots - this must be bogus.
> >
> > According to the theory, x^3 = 1 must have 3 roots (real+complex) - so
> > what am I missing ?
> >
> > regards
> > srp
>
> But i^4 = 1. So u're infinitely many roots actually reduce to just 3.
Also x^3 - 1 = 0 but x^3 - 1 factors into (x-1)(x^2 + x + 1), so that
either x-1 = 0 or x^2 + x + 1 = 0, and the latter is a quadratic
equation with two (complex) roots.
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