Sum of Two 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 $n \in \Z_{\ge 0}$ be a positive integer.


Then:

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


Proof

\(\displaystyle a^{2 n + 1} + b^{2 n + 1}\) \(=\) \(\displaystyle a^{2 n + 1} - \paren {-\paren {b^{2 n + 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle a^{2 n + 1} - \paren {-b}^{2 n + 1}\) as $n$ is odd
\(\displaystyle \) \(=\) \(\displaystyle \paren {a - \paren {-b} } \sum_{j \mathop = 0}^{2 n} a^{2 n - j} \paren {-b}^j\) Difference of Two Powers
\(\displaystyle \) \(=\) \(\displaystyle \paren {a + b} \sum_{j \mathop = 0}^{2 n} \paren {-1}^j a^{2 n - j} b^j\) simplifying

$\blacksquare$


Examples

Sum of Two Cubes

$x^3 + y^3 = \paren {x + y} \paren {x^2 - x y + y^2}$


Sum of Two Fifth Powers

$x^5 + y^5 = \paren {x + y} \paren {x^4 - x^3 y + x^2 y^2 - x y^3 + y^4}$


Also see


Sources