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

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

Mathematics
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Factor and simplify are different things.
(a-b)^4n
\[a^{4n}-b^{4n}\]\[=(a^{2n})^{2}-(b^{2n})^{2}=...\]

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Other answers:

If they share something, it can be factored.
I'm sure if you expand (a-b)^4n, you won't get a^4n - b^4n
okay yeah that was a stupid question :P I understand it now! thanks anyways guys!
\[\large{\color{blue}{a^{4n}-b^{4n} = (a^{2n})^2 - (b^{2n})^2 = (a^{2n} + b^{2n} )(a^{2n}-b^{2n})}}\]
a^2n - b^2n can be further factorised!
The identity used here is \(\large{\color{red}{a^2-b^2=(a+b)(a-b)}}\)
\[\large{(a^{2n}+b^{2n})(a^{2n}-b^{2n}) = (a^{2n}+b^{2n})[(a^n)^2 - (b^n)^2] \\= (a^{2n}+b^{2n}) (a^n + b^n)(a^n-b^n)}\]
but a^(2n)+b^(2n) cant be furthur factorized?
Yes @tok1997 a^(2n)+b^(2n) can not be further factorised but a^(2n)-b^(2n) can be.
Yup! unless... using complex number...
yeah ... @RolyPoly
thanks guys :)
You're welcome @tok1997

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