Power Series Expansion of Cube Root of 1 + x

Theorem
Let $x \in \R$ such that $-1 < x \le 1$.

Then:
 * $\sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots$