$${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$$

Let $$f(t) = {t^3} + {\left( {{\pi \over 2} - t} \right)^3}$$

Where $$t = {\tan ^{ - 1}}x$$ ; $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$

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

$$f(t) = {{3\pi } \over 2}{t^2} - {{3{\pi ^2}} \over 4}\,.\,t + {{{\pi ^3}} \over 8}$$

This is a quadratic equation of t.

Here, coefficient of t² term is $${{3\pi } \over 2}$$ which is > 0.

$$\therefore$$ It is a upward parabola.

Now, $$f'(t) = 3\pi t - {{3{\pi ^2}} \over 4}$$

$$f''(t) = 3\pi > 0$$

$$\therefore$$ $$3\pi t - {{3{\pi ^2}} \over 4} = 0$$

$$ \Rightarrow t = {\pi \over 4}$$ (minima)

$$\therefore$$ vertex of graph at $${\pi \over 4}$$

$$\therefore$$ Minimum value at $${\pi \over 4}$$ and maximum value at $$-$$$${\pi \over 2}$$.

$$\therefore$$ $$f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$$

$$f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$$

$$ = {{7{\pi ^3}} \over 8}$$

$$\therefore$$ $$k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$$

$$ \Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$$