Question

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

Answer 1

Factor and simplify are different things.

Answer 2

(a-b)^4n

Answer 3

\[a^{4n}-b^{4n}\]\[=(a^{2n})^{2}-(b^{2n})^{2}=...\]

Answer 4

If they share something, it can be factored.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

yeah ... @RolyPoly

Answer 15

thanks guys :)

Answer 16

You're welcome @tok1997