anonymous
  • anonymous
how can i factor: a^(4n)-b^(4n) Seems to me its already in its simplest form
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Factor and simplify are different things.
EpikRebel
  • EpikRebel
(a-b)^4n
anonymous
  • anonymous
\[a^{4n}-b^{4n}\]\[=(a^{2n})^{2}-(b^{2n})^{2}=...\]

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

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