Sum of Two Odd Powers/Examples/Sum of Two Cubes

Theorem

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

$\ds a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$

Let $x = a$ and $y = -b$.

Then:

$\ds x^3 + y^3$

$=$

$\ds x^3 - \paren {-y^3}$

$\ds $

$=$

$\ds x^3 - \paren {-y}^3$

$\ds $

$=$

$\ds \paren {x - \paren {-y} } \sum_{j \mathop = 0}^2 x^{n - j - 1} \paren {-y}^j$

$\ds $

$=$

$\ds \paren {x + y} \paren {x^2 + x \paren {-y} + \paren {-y}^2}$

$\ds $

$=$

$\ds \paren {x + y} \paren {x^2 - x y + y^2}$

$\blacksquare$

$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$

We have that $3 = 2 \times 1 + 1$.

Hence setting $n = 1$ gives the result.

$\blacksquare$