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Thread Subject:
ALGEBRA PROBLEM

Subject: ALGEBRA PROBLEM

From: vasu rao

Date: 19 May, 2011 14:15:04

Message: 1 of 9

FIND THE FACTORS FOR

a^6 - b^6

Subject: ALGEBRA PROBLEM

From: Steven_Lord

Date: 19 May, 2011 14:17:56

Message: 2 of 9



"vasu rao" <srirao17@gmail.com> wrote in message
news:ir38l8$sfa$1@newscl01ah.mathworks.com...
> FIND THE FACTORS FOR
>
> a^6 - b^6

Okay, done.

Now why don't you show me what your answer was?

--
Steve Lord
slord@mathworks.com
To contact Technical Support use the Contact Us link on
http://www.mathworks.com

Subject: ALGEBRA PROBLEM

From: Greg Heath

Date: 19 May, 2011 17:53:11

Message: 3 of 9

On May 19, 10:15 am, "vasu rao" <srira...@gmail.com> wrote:
> FIND THE FACTORS FOR
>
> a^6 - b^6

Two are obvious. Divide them out and you only have to find the
factors of a 4th order polynomial.

Piece of cake.

Greg

Subject: ALGEBRA PROBLEM

From: Florin Neacsu

Date: 19 May, 2011 18:30:21

Message: 4 of 9

"vasu rao" wrote in message <ir38l8$sfa$1@newscl01ah.mathworks.com>...
> FIND THE FACTORS FOR
>
> a^6 - b^6

Hi,

Disclaimer : I am not a native speaker so I might answer off the topic. If so, sorry...

Is this a homework? If so, I am quite impressed. We used to do this when we were in 5th grade so I am amazed you had the good idea to come and ask here.

Anyway :

a^6-b^6=(a^3)^2-(b^3)^2
            = (a^3-b^3)(a^3+b^3)
            = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2);

If you are also interested in a^6+b^6 (in some parts of the world called G. de Racquigny - Adanson) :

a^6+b^6 = (a^3-2ab^2)^3+(b^3-2a^2b)^2; (not a factorization but is a sum of squares)

Regards,
Florin

Subject: ALGEBRA PROBLEM

From: Nasser M. Abbasi

Date: 19 May, 2011 20:11:35

Message: 5 of 9

On 5/19/2011 11:30 AM, Florin Neacsu wrote:
> "vasu rao" wrote in message<ir38l8$sfa$1@newscl01ah.mathworks.com>...
>> FIND THE FACTORS FOR
>>
>> a^6 - b^6
>
> Hi,
>
> Disclaimer : I am not a native speaker so I might answer off the topic. If so, sorry...
>
> Is this a homework? If so, I am quite impressed. We used to do this when we were in 5th grade so I am amazed you had the good idea to come and ask here.
>
> Anyway :
>
> a^6-b^6=(a^3)^2-(b^3)^2
> = (a^3-b^3)(a^3+b^3)
> = (a-b)(a^2+ab+b^2)(a+b)(a^2-ab+b^2);
>
> If you are also interested in a^6+b^6 (in some parts of the world called G. de Racquigny - Adanson) :
>
> a^6+b^6 = (a^3-2ab^2)^3+(b^3-2a^2b)^2; (not a factorization but is a sum of squares)
>
> Regards,
> Florin

You are good. Your answer matches the computer answer: (matlab 2011a using syms)

EDU>> syms a b; factor(a^6 - b^6)
    
(a - b)*(a + b)*(a^2 + a*b + b^2)*(a^2 - a*b + b^2)


--Nasser

Subject: ALGEBRA PROBLEM

From: Roger Stafford

Date: 20 May, 2011 02:04:05

Message: 6 of 9

"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

Subject: ALGEBRA PROBLEM

From: Florin Neacsu

Date: 20 May, 2011 14:43:04

Message: 7 of 9

"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

Subject: ALGEBRA PROBLEM

From: Roger Stafford

Date: 20 May, 2011 19:39:05

Message: 8 of 9

"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

Subject: ALGEBRA PROBLEM

From: Florin Neacsu

Date: 20 May, 2011 22:33:04

Message: 9 of 9

"Roger Stafford" wrote in message <ir6g0p$do0$1@newscl01ah.mathworks.com>...
> "Florin Neacsu" wrote in message <ir5ulo$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

Hi,

I was giving an approach that was using basic calculus (pre high-school). Obviously your solution is more elegant and easily adaptable to a generalization of the question. And I do agree that a solution that spreads over more than one line can be beautiful.

Regards,
Florin

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