For skewness, expand the cubic and break up the sum. Factoring out constants (such as \(\mu\)) gives
\begin{align*}
& \frac{\sum_{k=1}^n ( x_k-\mu )^3}{n}\\
& = \frac{\sum_{k=1}^n x_k^3}{n} - 3 \mu \frac{\sum_{k=1}^n x_k^2 }{n} + 3 \mu^2 \frac{\sum_{k=1}^n x_k}{n} - \frac{\sum_{k=1}^n \mu^3}{n}\\
& = \frac{\sum_{k=1}^n x_k^3}{n} - 3 \mu (v + \mu^2) + 3 \mu^3 - \mu^3\\
& = \frac{\sum_{k=1}^n x_k^3}{n} - 3 \mu v - \mu^3\\
& = \frac{\sum_{k=1}^n x_k^3}{n} - 3 \mu \left ( \sum_{k=1}^n \frac{x_k^2}{n} - \mu^2 \right ) - \mu^3\\
& = \frac{\sum_{k=1}^n x_k^3}{n} - 3 \mu \sum_{k=1}^n \frac{x_k^2}{n} + 2\mu^2.
\end{align*}
and divide by the cube of the standard deviation to finish. Note that the first expansion in the derivation above can be used quickly if the data is collected in a table and powers easily computed.