- Thread starter
- #1

#### paulmdrdo

##### Active member

- May 13, 2013

- 386

how would i start factoring this

$m^8-n^8-2m^6n^2+2n^6m^2$

$m^8-n^8-2m^6n^2+2n^6m^2$

- Thread starter paulmdrdo
- Start date

- Thread starter
- #1

- May 13, 2013

- 386

how would i start factoring this

$m^8-n^8-2m^6n^2+2n^6m^2$

$m^8-n^8-2m^6n^2+2n^6m^2$

- Admin
- #2

You really need to post your attempts at these, this way we can see where you are at.

I would look at factoring the first two terms as the difference of squares, and the last two terms have a common factor as well (I would factor out $-2m^2n^2$)...what do you get?

- Thread starter
- #3

- May 13, 2013

- 386

this is where i can get to

$(m^4+n^4)(m^4-n^4)-2n^2m^2(m^4-n^4)$

$(m^2-n^2)(m^2+n^2)(m^4+n^4)-2n^2m^2(m^2-n^2)(m^2+n^2)$

$(m-n)(m+n)(m^2+n^2)(m^4+n^4)-2n^2m^2(m-n)(m+n)(m^2+n^2)$

- Admin
- #4

I would take this path:

\(\displaystyle \left(m^4+n^4 \right)\left(m^4-n^4 \right)-2m^2n^2\left(m^4-n^4 \right)\)

\(\displaystyle \left(m^4-n^4 \right)\left(m^4-2m^2n^2+n^4 \right)\)

Do you recognize that the second factor is a square of a binomial?

- Thread starter
- #5

- May 13, 2013

- 386

yes this is answer

$(m-n)^{3}(m+n)^{3}(m^2+n^2)$

- Admin
- #6

Yes, good job!yes this is answer

$(m-n)^{3}(m+n)^{3}(m^2+n^2)$