Expanding x^n-a^n without Binomial Theroem ?

[gaganspidey]

1. The problem statement, all variables and given/known data

http://i46.tinypic.com/33ddulu.jpg

This is the given Theorem in my book, everything seems fine except that I cannot figure how they expanded (xn - an)

2. Relevant equations

The Binomial Theorem

3. The attempt at a solution

According to me (xn - an) = {[(x+a)-a]n - an} and expanding it would yield terms containing the nC0,nC1 etc. but they haven't shown anything like this where did all this disappear ? Plus I know that (x-a) would come out common and get cancel by (x-a) in the denominator.

[vela]

The very first phrase in the proof tells you: Dividing (xn-an) by (x-a).

[gaganspidey]

^ That went over my head :redface:
I mean thats the question I am asking, how do I divide them ?

[vela]

Wikipedia has an example, but there are probably better sources that cover polynomial division. Synthetic division is probably the method you learned back in algebra.

http://en.wikipedia.org/wiki/Polynomial_long_division
http://en.wikipedia.org/wiki/Synthetic_division

[HallsofIvy]

Part of the confusion may be that they are NOT expanding- they are factoring which is, basically, the opposite of "expanding".
You probably already know the second degree version of that: x^2- y^2= (x- y)(x+ y).

The third degree version is x^3- y^3= (x- y)(x^2+ xy+ y^2).

In general x^n- y^n= x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1}

[gaganspidey]

Thanks very much vela & HallsofIvy, finally I get the hang of this thing, how clumsy of me not to think of it.

[Gurudev MJ]

There is small change required in the formula you mentioned.

(x^n - a^n) must be expanded in general as below.

(x^n - a^n) = (x - a) ( Rest of what you mentioned above after = sign).

So, now, (x - a) can be cancelled with the denominator in the problem raised above.

Suggest me if I am wrong.

[Mark44]

There is small change required in the formula you mentioned.

(x^n - a^n) must be expanded in general as below.
As already noted in this thread, xn - an is NOT being expanded; it is being factored.


(x^n - a^n) = (x - a) ( Rest of what you mentioned above after = sign).

So, now, (x - a) can be cancelled with the denominator in the problem raised above.

Suggest me if I am wrong.