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

It must be done 2 ways: A) Treating it as adifferent of squaresand B) Treating it as adifference of cubes

I'm really confused. Thanks for help!!

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- Oct 13th 2009, 07:58 PMMATHDUDE2How 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!! - Oct 13th 2009, 08:03 PMmr fantastic
- Oct 13th 2009, 08:20 PMMATHDUDE2
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) $\displaystyle (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)$

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

*Answers from the back of the book - Oct 13th 2009, 08:27 PMmr fantastic
- Oct 13th 2009, 08:42 PMMATHDUDE2
For A) I understand why would factor $\displaystyle (x-y)$ and $\displaystyle (x+y)$ but where is the $\displaystyle (x^2+xy+y^2)(x^2-xy+y^2)$ coming from?

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