how can i factor:
a^(4n)-b^(4n)
Seems to me its already in its simplest form

Question

how can i factor: 
a^(4n)-b^(4n)
Seems to me its already in its simplest form

anonymous · Answer

Factor and simplify are different things.

epikrebel · Answer

(a-b)^4n

anonymous · Answer

$$a^{4n}-b^{4n}$$$$=(a^{2n})^{2}-(b^{2n})^{2}=...$$

epikrebel · Answer

If they share something, it can be factored.

anonymous · Answer

I'm sure if you expand (a-b)^4n, you won't get a^4n - b^4n

anonymous · Answer

okay yeah that was a stupid question :P I understand it now! 
thanks anyways guys!

anonymous · Answer

$$\large{\color{blue}{a^{4n}-b^{4n} = (a^{2n})^2 - (b^{2n})^2 = (a^{2n} + b^{2n} )(a^{2n}-b^{2n})}}$$

anonymous · Answer

a^2n - b^2n can be further factorised!

anonymous · Answer

The identity used here is $\large{\color{red}{a^2-b^2=(a+b)(a-b)}}$

anonymous · Answer

$$\large{(a^{2n}+b^{2n})(a^{2n}-b^{2n}) = (a^{2n}+b^{2n})[(a^n)^2 - (b^n)^2] \= (a^{2n}+b^{2n}) (a^n + b^n)(a^n-b^n)}$$

anonymous · Answer

but a^(2n)+b^(2n) cant be furthur factorized?

anonymous · Answer

Yes @tok1997 a^(2n)+b^(2n) can not be further factorised but a^(2n)-b^(2n) can be.

anonymous · Answer

Yup! unless... using complex number...

anonymous · Answer

yeah ... @RolyPoly

anonymous · Answer

thanks guys :)

anonymous · Answer

You're welcome @tok1997