Sum of Two Odd Powers/Examples/Sum of Two Cubes/Proof 2

Proof
From Sum of Two Odd Powers:


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

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

Hence setting $n = 1$ gives the result.