Power Series Expansion of Reciprocal of Cube Root of 1 + x

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

Then:


 * $\dfrac 1 {\sqrt [3] {1 + x} } = 1 - \dfrac 1 3 x + \dfrac {1 \times 4} {3 \times 6} x^2 - \dfrac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots$