Is There Any Inverse Trigonometric Identity?

Let $\arcsin x = \alpha$ and $\arccos x = \beta$ , then we have:

$\begin{cases} \sin \alpha = x & \Rightarrow \cos \alpha = \sqrt{1-x^2} \\ \cos \beta = x & \Rightarrow \sin \beta = \sqrt{1-x^2} \end{cases}$

$\begin{aligned} \Rightarrow \sin (\alpha + \beta) & = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ & = x^2 + 1 - x^2 \\ & = 1 \\ \Rightarrow \alpha + \beta & = \frac{\pi}{2} \\ \Rightarrow \beta & = \frac{\pi}{2} - \alpha \end{aligned}$

Now, we have:

$\begin{aligned} f(x) & = \alpha^3 + \beta^3 \\ & = \alpha^3 + \left( \frac{\pi}{2} - \alpha\right)^3 \\ & = \alpha^3 + \left(\frac{\pi^3}{8} - \frac{3\pi^2\alpha}{4} + \frac{3\pi \alpha^2}{2} - \alpha^3 \right) \\ & = \frac{\pi}{8} \left(12\alpha^2 - 6\pi \alpha + \pi^2 \right) \\ & = \frac{12\pi}{8} \left(\alpha^2 - \frac{\pi \alpha}{2} + \frac{\pi^2}{12} \right) \\ & = \frac{3\pi}{2} \left( \left[\alpha - \frac{\pi}{4}\right]^2 + \frac{\pi^2}{48} \right) \quad \text{for } \alpha = \arcsin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}$

We note that $f(x) > 0$ and it is maximum when $\alpha = - \frac{\pi}{2}$ and minimum, $\alpha = \frac{\pi}{4}$ . Therefore, we have:

$\begin{aligned} f_{max}(x) & = \frac{3\pi}{2}\left( \left[- \frac{\pi}{2} - \frac{\pi}{4}\right]^2 + \frac{\pi^2}{48} \right) = \frac{3\pi}{2} \left(\frac{9\pi^2}{16} + \frac{\pi^2}{48} \right) = \frac{7\pi^3}{8} \\ f_{min}(x) & = \frac{3\pi}{2}\left( \left[\frac{\pi}{4} - \frac{\pi}{4}\right]^2 + \frac{\pi^2}{48} \right) = \frac{\pi^3}{32} \\ \Rightarrow f_{max}(x) + f_{min}(x) & = \frac{7\pi^3}{8} + \frac{\pi^3}{32} = \frac{29\pi^3}{32} \end{aligned}$

$\Rightarrow a + b + c = 29 + 3 + 32 = \boxed{64}$