\begin{align*} \gamma_1 = & \sum_{x=1}^n x^3 \cdot \frac{1}{n} - 3 \mu \sum_{x=1}^n x^2 \cdot \frac{1}{n} + 2\mu^3\\ & = \frac{n^2(n+1)^2}{4n} - 3\frac{(n(n+1)(2n+1))}{2n} \frac{1+n}{2} + 2 \left ( \frac{1+n}{2}\right )^3 \\ & = \frac{n^2(n+1)^2}{4n} - \frac{(n+1)^2 (n(2n+1)}{4n} + \frac{(n+1)^3}{4}\\ & = \frac{(n+1)^2}{4} \left [ n - 2n -1 + (n+1) \right ]\\ & = 0 \end{align*}