Difference of Two Even-Times Odd Powers

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.

$\blacksquare$