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:
\(\ds a^{2 n} - b^{2 n}\) | \(=\) | \(\ds \paren {a - b} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {a + b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } \paren {a^{n - 1} - a^{n - 2} b + a^{n - 3} b^2 - \dotsb - a b^{n - 2} + b^{n - 1} }\) |
Proof
\(\ds a^{2 n} - b^{2 n}\) | \(=\) | \(\ds \paren {a^n}^2 - \paren {b^n}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^n - b^n} \paren {a^n + b^n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a^n + b^n}\) | Difference of Two Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}\) | Sum of Two Odd Powers |
whence the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.22$