${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3}$ 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 t 2 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}. $\therefore$ $f\left( {{\pi \over 4}} \right) = {{{\pi ^3}} \over {64}} + {\left( {{\pi \over 2} - {\pi \over 4}} \right)^3} = {{{\pi ^3}} \over {32}}$ $f\left( { - {\pi \over 2}} \right) = - {{{\pi ^3}} \over 8} + {\pi ^3}$ $= {{7{\pi ^3}} \over 8}$ $\therefore$ $k{\pi ^3} \in \left[ {{{{\pi ^3}} \over {32}},\,{{7{\pi ^3}} \over 8}} \right)$ $\Rightarrow k \in \left[ {{1 \over {32}},\,{7 \over 8}} \right)$