# Difference of Two Powers/Proof 3

## Theorem

Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.

Let $n \in \N$ such that $n \ge 2$.

Then for all $a, b \in \mathbb F$:

 $\ds a^n - b^n$ $=$ $\ds \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ $\ds$ $=$ $\ds \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }$

## Proof

 $\ds \sum_{j \mathop = 0}^{n - 1} x^j$ $=$ $\ds \frac {x^n - 1} {x - 1}$ $\ds \leadsto \ \$ $\ds \paren {\dfrac a b}^n - 1$ $=$ $\ds \paren {\dfrac a b - 1} \sum_{j \mathop = 0}^{n - 1} \paren {\dfrac a b}^j$ $\ds \leadsto \ \$ $\ds \paren {\dfrac {a^n - b^n} {b^n} }$ $=$ $\ds \paren {\dfrac {a - b} b} \paren {\paren {\dfrac a b}^{n - 1} + \paren {\dfrac a b}^{n - 2} + \dotsb + \paren {\dfrac a b}^1 + 1}$ $\ds \leadsto \ \$ $\ds a^n - b^n$ $=$ $\ds \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + \dotsb + a b^{n - 2} + b^{n - 1} }$ multiplying both sides by $b^n$

$\blacksquare$