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Subject: Re: ALGEBRA PROBLEM
Date: Fri, 20 May 2011 19:39:05 +0000 (UTC)
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"Florin Neacsu" wrote in message <ir5ulo$idc$1@newscl01ah.mathworks.com>...
> "Roger Stafford" wrote in message <ir4i6l$ppu$1@newscl01ah.mathworks.com>...
> > "Florin Neacsu" wrote in message <ir3njt$f4h$1@newscl01ah.mathworks.com>...
> > >             = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2);
> > - - - - - - - - - - -
> >   Why not take it all the way to six factors?
> > 
> > Roger Stafford
> 
> Hi,
>  Something like :
> 
>        = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2);
>        = (a-b)(a^2+2ab+b^2 -ab)(a+b)(a^2-2ab+b^2+ab)
>        = (a-b)((a+b)^2 - sqrt(ab)^2)(a+b)((a-b)^2+sqrt(ab)^2)
>        = (a-b)((a+b-sqrt(ab))(a+b+sqrt(ab))(a+b)(a-b+sqrt(ab))(a-b-sqrt(ab)) 
> 
> Six factors, but it looks "uglier" to me.  
> Regards,
> Florin
- - - - - - - - -
  Well, what I had in mind were factors linear in a and b using in this case the sixth roots of unity:

 a^6-b^6 = (a-b*exp(0/3*pi*i)) * (a-b*exp(1/3*pi*i)) * ...
           (a-b*exp(2/3*pi*i)) * (a-b*exp(3/3*pi*i)) * ...
           (a-b*exp(4/3*pi*i)) * (a-b*exp(5/3*pi*i))

which is actually just your answer with the two quadratic factors divided further into four linear factors.  Of course that brings one into the complex world but some mathematicians would regard it as beautiful even if it stretches for three lines.

Roger Stafford