SOLUTION: factor completely a^6n - b^6n

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Question 39750: factor completely
a^6n - b^6n
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Factor completely:
a%5E%286n%29+-+b%5E%286n%29 You should recognise this as the difference of two squares:
%28a%5E%283n%29%29%5E2+-+%28b%5E%283n%29%29%5E2
The difference of two squares is factorable as follows:
A%5E2+-+B%5E2+=+%28A+-+B%29%28A+%2B+B%29
Applying this to your problem, the A+=+a%5E%283n%29 and the B+=+b%5E%283n%29, so we have:

But we're not quite finished because you may also recognise that the two factors that we now have are the difference and the sum of two cubes, which are both factorable as follows:
A%5E3+-+B%5E3+=+%28A+-+B%29%28A%5E2+%2B+AB+%2B+B%5E2%29 and
A%5E3+%2B+B%5E3+=+%28A+%2B+B%29%28A%5E2+-+AB+%2B+B%5E2%29
So, applying this to the two factors we found from the difference of two squares, we get:
...and you could rearrange this a little:
The end!