Let $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$ be a real-valued function defined on its domain. Then the sum of the greatest and the least values of $f(x)$ is

Given $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$ with domain $x \in [-1, 1]$.

Using the identity $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$, we can write:

$f(x) = (\sin^{-1} x)^2 + \left(\dfrac{\pi}{2} - \sin^{-1} x\right)^2$

Let $\sin^{-1} x = t$, where $t \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

$f(t) = t^2 + \left(\dfrac{\pi}{2} - t\right)^2 = 2t^2 - \pi t + \dfrac{\pi^2}{4}$

Completing the square, we get:

$f(t) = 2\left(t^2 - \dfrac{\pi}{2}t\right) + \dfrac{\pi^2}{4} = 2\left(t - \dfrac{\pi}{4}\right)^2 - 2\left(\dfrac{\pi^2}{16}\right) + \dfrac{\pi^2}{4}$

$f(t) = 2\left(t - \dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{8}$

The minimum value of $f(t)$ occurs at $t = \dfrac{\pi}{4}$, which lies in the interval $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Minimum value $= \dfrac{\pi^2}{8}$

The maximum value of $f(t)$ occurs at the endpoint furthest from $t = \dfrac{\pi}{4}$, which is $t = -\dfrac{\pi}{2}$.

Maximum value $= 2\left(-\dfrac{\pi}{2} - \dfrac{\pi}{4}\right)^2 + \dfrac{\pi^2}{8} = 2\left(-\dfrac{3\pi}{4}\right)^2 + \dfrac{\pi^2}{8} = \dfrac{9\pi^2}{8} + \dfrac{\pi^2}{8} = \dfrac{5\pi^2}{4}$

The sum of the greatest and the least values of $f(x)$ is:

$\dfrac{5\pi^2}{4} + \dfrac{\pi^2}{8} = \dfrac{10\pi^2}{8} + \dfrac{\pi^2}{8} = \dfrac{11\pi^2}{8}$

Answer: $\dfrac{11\pi^2}{8}$