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Factoring: Difference of Cubes


Date: 09/05/2001 at 03:19:37
From: brian park
Subject: Factoring

I'm wondering if you could help me solve this problem:

Factor a^3b^3 - 8x^6y^9

Please help me. Thank you.


Date: 09/05/2001 at 11:44:41
From: Doctor Jubal
Subject: Re: Factoring

Hi Brian,

Thanks for writing to Dr. Math.

Solving this problem relies on a common formula for factoring the 
difference of two cubes:

  x^3 - y^3 = (x-y)(x^2 + xy + y^2)

This is just an extension of the the formula for the difference of two
squares that you've probabaly seen:

  x^2 - y^2 = (x-y)(x+y)

In fact, the formula can be extended to the difference between any two 
like powers:

  x^4 - y^4 = (x-y)(x^3 + x^2*y + xy^2 + y^3)
  x^5 - y^5 = (x-y)(x^4 + x^3*y + x^2*y^2 + xy^3 + y^4)
  x^n - y^n = (x-y)(x^(n-1) + x^(n-2)y + ... + xy^(n-2) + y^(n-1))

Do you see why this works? The x and the y in the first factor have
opposite signs, and when you multiply them by all the terms in the 
second factor, all the terms exist in pairs with opposite signs and 
cancel out, except the x^n and y^n terms.

To demonstrate this for the fourth powers:

  (x-y)(x^3 + x^2*y + xy^2 + y+3) = 

  (x^4 + x^3*y + x^2*y^2 + xy^3) -
        (x^3*y + x^2*y^2 + xy^3 + y^4) = 

  (x^4 - y^4)

Now, back to your original question:

Do you see that both a^3b^3 and 8x^6y^9 are cubes? Every factor in 
each of them is raised to the third power.  a^3b^3 is (ab)^3, and 
8x^6y^9 is (2x^2*y^3)^3.

So, you can use the formula for the difference of two cubes to factor 
it:

  (a^3b^3 - 8x^6y^9) = 
    (ab - 2x^2*y^3)((ab)^2 + (ab)(2x^2*y^3) + (2x^2*y^3)^2)

I'll leave it to you to do any final simplification you might want to 
do with the final answer, but that's the gist of it.

Does this help? Write back if you'd like to talk about this some more, 
or if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents

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