\begin{align*} \mu & = E[X] = \int_a^b x \frac{1}{b-a} dx\\ & = \frac{x^2}{2(b-a)} right |_a^b\\ & = \frac{b^2-a^2}{2(b-a)} = \frac{b+a}{2}. \end{align*}

\begin{align*} \sigma^2 & = E[X^2] - \mu^2 = \int_a^b x^2 \frac{1}{b-a} dx - \mu^2\\ & = \frac{x^3}{3(b-a)} right |_a^b - \left ( \frac{a+b}{2} \right )^2\\ & = \frac{b^3-a^3}{3(b-a)} - \frac{a^2 + 2ab + b^2}{4}\\ & = \frac{4 b^2 + 4 ab + 4 a^2 - 3a^2 - 6 ab - 3b^2}{12}\\ & = \frac{b^2-2ab+a^2}{12} = \frac{(b-a)^2}{12}. \end{align*}

\begin{align*} \gamma_0 & = E[X^3] - 3 \mu E[X^2] + 2 \mu^3\\ & = \int_a^b x^3 \frac{1}{b-a} dx - 3 \mu \frac{b^3-a^3}{3(b-a)} + 2 \left ( \frac{a+b}{2} \right )^3\\ & = \frac{x^4}{4(b-a)} right |_a^b - 3 \frac{a+b}{2} \cdot \frac{b^3-a^3}{3(b-a)} + 2 \frac{a^3 + 3a^2 b + 3a b^2 + b^3}{8} \\ & = \text{a miracle of algebra}\\ & = 0. \end{align*}