# How to solve Difference of Cubes?

• October 13th 2009, 07:58 PM
MATHDUDE2
How to solve Difference of Cubes?
Hi, can someone show me how to solve the expression x^6 - y^6?

It must be done 2 ways: A) Treating it as a different of squares and B) Treating it as a difference of cubes

I'm really confused. Thanks for help!!
• October 13th 2009, 08:03 PM
mr fantastic
Quote:

Originally Posted by MATHDUDE2
Hi, can someone show me how to solve the expression x^6 - y^6?

It must be done 2 ways: A) Treating it as a different of squares and B) Treating it as a difference of cubes

I'm really confused. Thanks for help!!

A) $x^6 - y^6 = (x^3)^2 - (y^3)^2$.

B) $x^6 - y^6 = (x^2)^3 - (y^2)^3$.

And I assume you're familar with the formulae for the difference of two squares and for the difference of two cubes.
• October 13th 2009, 08:20 PM
MATHDUDE2
Quote:

Originally Posted by mr fantastic
A) $x^6 - y^6 = (x^3)^2 - (y^3)^2$.

B) $x^6 - y^6 = (x^2)^3 - (y^2)^3$.

And I assume you're familar with the formulae for the difference of two squares and for the difference of two cubes.

Thanks for the response. And yes, I am familiar with the basic A-B formula. However I don't understand why it expands to become:

A) $(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$
B) $(x-y)(x+y)(x^4+x^2y^2+y^4)$ --> if this is a difference of squares, why is there a (x+y)?

*Answers from the back of the book
• October 13th 2009, 08:27 PM
mr fantastic
Quote:

Originally Posted by MATHDUDE2
Thanks for the response. And yes, I am familiar with the basic A-B formula. However I don't understand why it expands to become:

A) $(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$
B) $(x-y)(x+y)(x^4+x^2y^2+y^4)$ --> if this is a difference of squares, why is there a (x+y)?

*Answers from the back of the book

• October 13th 2009, 08:42 PM
MATHDUDE2
Quote:

Originally Posted by mr fantastic

For A) I understand why would factor $(x-y)$ and $(x+y)$ but where is the $(x^2+xy+y^2)(x^2-xy+y^2)$ coming from?

For B) I know where the $(x-y)$ is coming from (difference of cubes, [a-b]) but where is the $(x+y)$ and $(x^4+x^2y^2+y^4)$ coming from? I kind of understand the trinomial part given that $[a^2+ab+b^2]$. But other than that, I'm pretty much lost.
• October 13th 2009, 08:48 PM
mr fantastic
Quote:

Originally Posted by MATHDUDE2
For A) I understand why would factor $(x-y)$ and $(x+y)$ but where is the $(x^2+xy+y^2)(x^2-xy+y^2)$ coming from?

For B) I know where the $(x-y)$ is coming from (difference of cubes, [a-b]) but where is the $(x+y)$ and $(x^4+x^2y^2+y^4)$ coming from? I kind of understand the trinomial part given that $[a^2+ab+b^2]$. But other than that, I'm pretty much lost.

$A^3 - B^3 = ....$ and $A^3 + B^3 = ....$.

I say again, show all your working. (I've asked this for a very good reason so please do it if you want help).