$f(x)=\frac{x^3-1}{x^2+2}$ $x=2 \to =\frac{2^3-1}{2^2+2}=\frac{7}{6}$ $x=-2 \to =\frac{(-2)^3-1}{(-2)^2+2}=\frac{-8-1}{6}=\frac{-9}{6}$ $\frac{7}{6}-\frac{-9}{6}=\frac{7+9}{6}=\frac{16}{6}=\frac{8}{3}$