### Instruction: Let
\[f(x) = (\arccos x)^3 + (\arcsin x)^3.\]Find the range of $f(x).$  All functions are in radians.### Response: To find the range of a function, I need to think about what values it can possibly output for different inputs.
The inputs of $f(x)$ are values of $x$ that satisfy $-1 \leq x \leq 1$, since those are the domains of the arccosine and arcsine functions.
The outputs of $f(x)$ are values of $(\arccos x)^3 + (\arcsin x)^3$, which depend on the values of $\arccos x$ and $\arcsin x$.
Since $\arccos x$ and $\arcsin x$ are inverse functions of cosine and sine, respectively, I can use some trigonometric identities to relate them.
One identity that seems useful is $\arccos x + \arcsin x = \frac{\pi}{2}$, which holds for any $x$ in the domain.
This means that $(\arccos x)^3 + (\arcsin x)^3$ can be rewritten as
\[(\arccos x)^3 + (\arcsin x)^3 = (\arccos x)^3 + (\frac{\pi}{2} - \arccos x)^3.\]
Now I can see that this is a sum of two cubic functions of $\arccos x$, which is a monotonic function that ranges from $\pi$ to $0$ as $x$ ranges from $-1$ to $1$.
Therefore, I can find the minimum and maximum values of $f(x)$ by plugging in the endpoints of the range of $\arccos x$, namely $\pi$ and $0$.
When $\arccos x = \pi$, we have
\[(\arccos x)^3 + (\frac{\pi}{2} - \arccos x)^3 = \pi^3 + (-\frac{\pi}{2})^3 = \frac{7}{8}\pi^3.\]
When $\arccos x = 0$, we have
\[(\arccos x)^3 + (\frac{\pi}{2} - \arccos x)^3 = 0^3 + (\frac{\pi}{2})^3 = \frac{\pi^3}{8}.\]
Since $\arccos x$ decreases as $x$ increases, we can conclude that $f(x)$ also decreases as $x$ increases, and hence the minimum and maximum values of $f(x)$ are $\frac{\pi^3}{8}$ and $\frac{7}{8}\pi^3$, respectively.
Therefore, the range of $f(x)$ is the closed interval
\[\left[\frac{\pi^3}{8}, \frac{7}{8}\pi^3\right].\]