Difference of Two Powers/Proof 3

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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

From Sum of Geometric Sequence:


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

$\blacksquare$


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