Difference of Two Even-Times Odd Powers

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Theorem

Let $\F$ be one of the standard number systems, that is $\Z, \Q, \R$ and so on.

Let $n \in \Z_{> 0}$ be a (strictly) positive odd integer.


Then:

\(\displaystyle a^{2 n} - b^{2 n}\) \(=\) \(\displaystyle \paren {a - b} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {a + b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } \paren {a^{n - 1} - a^{n - 2} b + a^{n - 3} b^2 - \dotsb - a b^{n - 2} + b^{n - 1} }\)


Proof

\(\displaystyle a^{2 n} - b^{2 n}\) \(=\) \(\displaystyle \paren {a^n}^2 - \paren {b^n}^2\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a^n - b^n} \paren {a^n + b^n}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a^n + b^n}\) Difference of Two Powers
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}\) Sum of Two Odd Powers

whence the result.

$\blacksquare$


Sources