Thread: Factoring a large Diff of Squares

y = x^1000 - 1

I would like to factor out an x-1 from this but I don't remember which rules apply.

1) Do I apply something similar to a difference of squares and just keep going until I get (x-1)(....)?

(x^500)^2 - (1)^2 = (x^500+1)(x^500-1) = (x^500+1)(x^250+1)(x^250-1) = (x^500+1)(x^250+1)(x^125+1)(x^125-1) = now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent ... well I'll try difference of cubes in that last factor...: = (x^500+1)(x^250+1)(x^125+1)(x^5-1)(x^10+x^5+1)... Okay, now I'm stuck, unless I remember how to factor (x^10+x^5+1)...


2) Or do I use some kind of Binomial Theorem.

Is it true that
x^1000 - 1 = (x-1)(x^999+x^998+x^997......+1) ?


3) Do I have to remember Pascal's Triangle at all?

For example, if number 2 above is true, then do I have to put in pascal's coefficients? ie.
(x-1)(1x^999 + 999x^998 + ...... + 1)?

Hello, krystaline86!

This is an ugly problem . . .

Factor: .

1) Do I apply something similar to a difference of squares
and just keep going until I get: ?



. . .

. . . . . . . Good!

now I'm stuck and I couldn't even get it to x-1 because I ran into an odd exponent.
But odd exponents are usually "a good thing".

From what you wrote, I assume you are familiar with these formulas:

. .


So: .


Then: .


And: .


. . See?

Excellent. So my idea about using the Binomial Theorem was right then?

The answer would be:

(x^1000 - 1) = (x-1)(x^999+x^998+x^997+...+x+1)

Right?

I just want to know what that second factor expanded out would look like in the form of how I have it there.

OH, and how do I express that properly? Can I use a factorial somehow?