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

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Homework Statement



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

Homework Equations



The Binomial Theorem

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.
 

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The very first phrase in the proof tells you: Dividing (xn-an) by (x-a).
 
^ That went over my head :redface:
I mean that's the question I am asking, how do I divide them ?
 
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: [itex]x^2- y^2= (x- y)(x+ y)[/itex].

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

In general [itex]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}[/itex]
 
Thanks very much vela & HallsofIvy, finally I get the hang of this thing, how clumsy of me not to think of it.
 
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 canceled with the denominator in the problem raised above.

Suggest me if I am wrong.
 
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Gurudev MJ said:
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.
Gurudev MJ said:
(x^n - a^n) = (x - a) ( Rest of what you mentioned above after = sign).

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

Suggest me if I am wrong.