Hi there - I took some notes in my math class but I'm not sure I wrote them down correctly. Could someone please verify if this is correct or not?

Concerning the factors of (x^n)-(a^n):
When f(x)=(x^n)-(a^n) and f(x-a)=0, (x-a) is a root.
For n=2 it can be shown by long division that (x^2)-(a^2)=(x-a)(x-n).
For n=3: (x^3)-(a^3)=[(x-a)((x^2)-ax-(a^2))]
In general: (x^n)-(a^n)=[(x-a)((x^(n-1))+a(x^(n-2))+...+(a^(n-1))x]

Question

Hi there - I took some notes in my math class but I'm not sure I wrote them down correctly. Could someone please verify if this is correct or not?

Concerning the factors of (x^n)-(a^n):
When f(x)=(x^n)-(a^n) and f(x-a)=0, (x-a) is a root. 
For n=2 it can be shown by long division that (x^2)-(a^2)=(x-a)(x-n).
For n=3: (x^3)-(a^3)=[(x-a)((x^2)-ax-(a^2))]
In general: (x^n)-(a^n)=[(x-a)((x^(n-1))+a(x^(n-2))+...+(a^(n-1))x]

nikolas · Answer

It's the right side of those last two equations that I'm most concerned about as they don't seem to match each other.

nikolas · Answer

Right, I think I've got it, it's just a difference of two powers... so:
For n=2: (x^2)-(a^2)=(x+a)(x-a)
For n=3: (x^3)-(a^3)=[(x-a)((x^2)+ax+(a^2))]
In general: (x^n)-(a^n)=[(x-a)((x^(n-1))+a(x^(n-2))+...+(a^(n-1))]

Key thing to realize for me was that the degree of each term is always equal to n-1.