# Power Series Expansion of Cube Root of 1 + x

## Theorem

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

Then:

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

## Proof

 $\ds \sqrt [3] {1 + x}$ $=$ $\ds \paren {1 + x}^{\frac 1 3}$ $\ds$ $=$ $\ds \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 3}^{\underline k} } {k!} x^k$ General Binomial Theorem $\ds$ $=$ $\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 3 - j} } {k!} x^k$ Definition of Falling Factorial and extracting $k = 0$ $\ds$ $=$ $\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 3 j} } {3^k k!} x^k$ simplifying $\ds$ $=$ $\ds \sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots$

$\blacksquare$