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.
For kurtosis, similarly expand the quartic and break up the sum as before. Then,
\begin{align*}
& \frac{\sum_{k=1}^n ( x_k-\mu )^4}{n}\\
& = \frac{\sum_{k=1}^n x_k^4}{n} - 4 \mu \frac{\sum_{k=1}^n x_k^3 }{n} + 6 \mu^2 \frac{\sum_{k=1}^n x_k^2}{n} - 4 \mu^3 \frac{\sum_{k=1}^n x_k}{n} + \frac{\sum_{k=1}^n \mu^4}{n}\\
& = \frac{\sum_{k=1}^n x_k^4}{n} - 4 \mu \frac{\sum_{k=1}^n x_k^3 }{n} + 6 \mu^2 (v + \mu^2) - 4 \mu^4 + \mu^4\\
& = \frac{\sum_{k=1}^n x_k^4}{n} - 4 \mu \frac{\sum_{k=1}^n x_k^3 }{n} + 6 \mu^2 v - 3 \mu^4
\end{align*}
and then divide by the fourth power of the standard deviation. Note again that the first expansion in the derivation above might also be a useful shortcut.