# Homework Help: Expanding x^n-a^n without Binomial Theroem ?

1. Jun 30, 2010

### gaganspidey

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

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.

2. Jun 30, 2010

### vela

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

3. Jun 30, 2010

### gaganspidey

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

4. Jun 30, 2010

### vela

Staff Emeritus
5. Jun 30, 2010

### 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 $x^n- y^n= x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1}$

6. Jun 30, 2010

### gaganspidey

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

7. Oct 13, 2011

### 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.

8. Oct 13, 2011

### Staff: Mentor

As already noted in this thread, xn - an is NOT being expanded; it is being factored.