$$\begin{aligned} & \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\ & =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\ & =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c \end{aligned}$$

So $g(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x$

$$\begin{aligned} & g\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\ & g^{\prime}(x)=x^3 \sin x \\ & g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\ & 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48 \end{aligned}$$

So $\alpha+\beta-\gamma=55$