The set of all values of $$k$$ for which $$(\tan^{-1}x)^3 + (\cot^{-1}x)^3 = k\pi^3, x \in R$$, is the interval

We need to find all values of $$k$$ such that $$(\tan^{-1}x)^3 + (\cot^{-1}x)^3 = k\pi^3$$ has a solution for some $$x \in \mathbb{R}$$.

Let $$t = \tan^{-1}x$$. Then $$\cot^{-1}x = \frac{\pi}{2} - t$$, where $$t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

The equation becomes:

$$ t^3 + \left(\frac{\pi}{2} - t\right)^3 = k\pi^3 $$

Expand using the identity $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$ where $$a = t$$ and $$b = \frac{\pi}{2} - t$$:

$$a + b = \frac{\pi}{2}$$

$$ a^2 - ab + b^2 = (a+b)^2 - 3ab = \frac{\pi^2}{4} - 3t\left(\frac{\pi}{2} - t\right) $$

$$ = \frac{\pi^2}{4} - \frac{3\pi t}{2} + 3t^2 $$

So:

$$ k\pi^3 = \frac{\pi}{2}\left(\frac{\pi^2}{4} - \frac{3\pi t}{2} + 3t^2\right) $$

$$ k\pi^3 = \frac{\pi^3}{8} - \frac{3\pi^2 t}{4} + \frac{3\pi t^2}{2} $$

Dividing by $$\pi^3$$:

$$ k = \frac{1}{8} - \frac{3t}{4\pi} + \frac{3t^2}{2\pi^2} $$

Let $$f(t) = \frac{3t^2}{2\pi^2} - \frac{3t}{4\pi} + \frac{1}{8}$$. This is a quadratic in $$t$$ opening upward.

The minimum occurs at:

$$ t = \frac{\frac{3}{4\pi}}{2 \cdot \frac{3}{2\pi^2}} = \frac{\frac{3}{4\pi}}{\frac{3}{\pi^2}} = \frac{3}{4\pi} \cdot \frac{\pi^2}{3} = \frac{\pi}{4} $$

Minimum value of $$k$$:

$$ k_{min} = \frac{3}{2\pi^2}\cdot\frac{\pi^2}{16} - \frac{3}{4\pi}\cdot\frac{\pi}{4} + \frac{1}{8} = \frac{3}{32} - \frac{3}{16} + \frac{1}{8} = \frac{3}{32} - \frac{6}{32} + \frac{4}{32} = \frac{1}{32} $$

Find the range. Since $$t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ and $$f(t)$$ is a parabola opening upward with minimum at $$t = \frac{\pi}{4}$$:

At $$t = \frac{\pi}{4}$$: $$k = \frac{1}{32}$$ (achieved when $$x = \tan\frac{\pi}{4} = 1$$).

As $$t \to -\frac{\pi}{2}$$:

$$ k \to \frac{3}{2\pi^2}\cdot\frac{\pi^2}{4} + \frac{3}{4\pi}\cdot\frac{\pi}{2} + \frac{1}{8} = \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{7}{8} $$

Since $$t = -\frac{\pi}{2}$$ is not attained (it would require $$x \to -\infty$$), the upper bound $$\frac{7}{8}$$ is not achieved.

At $$t = \frac{\pi}{4}$$ (i.e., $$x = 1$$), the minimum $$\frac{1}{32}$$ is achieved.

Therefore, the set of all values of $$k$$ is $$\left[\frac{1}{32}, \frac{7}{8}\right)$$.

The answer is Option A: $$\left[\dfrac{1}{32}, \dfrac{7}{8}\right)$$.