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### Factor Completely

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Date: 02/13/98 at 13:57:03
From: Anonymous
Subject: Algebra - factor completely

The problems are x to the twelth power minus y to the twelth power,
And x to the twelth power plus y to the twelth power:

x^12-y^12    and   x^12+y^12

The problems look so scary I can't move past pulling out the x-y in
the first problem and the x+y in the second.

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Date: 02/16/98 at 12:48:27
From: Doctor Wolf
Subject: Re: Algebra - factor completely

Hi,

No doubt about it ... these are two big, ugly factoring problems. I'll
work on the first one and leave the second one for you.

Now x^12 - y^12 could be considered a "difference of two squares"
since x^12 = (x^6)^2 and y^12 = (y^6)^2.

Or x^12 - y^12 could be considered a "difference of two cubes"
since x^12 = (x^4)^3, and likewise for y.

There are standard forms for factoring a difference of two squares,
and for both the sum and difference of two cubes. Either way you
approach it, you need to factor until no more can (seemingly) be done
... so here goes!

x^12 - y^12 = (x^6 + y^6)(x^6 - y^6)  as a diff. of 2 squares
and
x^6 + y^6  = (x^2 + y^2)(x^4 - x^2*y^2 + y^4) as a sum of 2 cubes.

Neither quantity immediately above can be factored any further ...

While x^6 - y^6  = (x^3 + y^3)(x^3 - y^3) as a diff. of 2 squares.

But  x^3 + y^3  = (x + y)(x^2 -xy + y^2) as a sum of 2 cubes,
and  x^3 - y^3  = (x - y)(x^2 +xy + y^2) as a diff. of 2 cubes.

So finally ... x^12 - y^12 =

(x^2 + y^2)(x^4 - x^2*y^2 + y^4)(x+y)(x^2 -xy + y^2)(x-y)(x^2 +xy +
y^2).

Why don't you begin the second problem by considering it a sum of two
cubes:

x^12 + y^12 = (x^4)^3 + (y^4)^3

Good luck, and I'd like to know how you came out!

-Doctor Wolf,  The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra

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