Maximizing and minimizing inverse trigo functions

Let $\begin{cases} \alpha = \sin^{-1} x & \implies \sin \alpha = x \\ \beta = \cos^{-1} x & \implies \cos \beta = x \end{cases}$

$\begin{aligned} \implies x = \sin \alpha & = \cos \beta \\ & = \sin \left(\frac \pi2 - \beta \right) \\ \implies \alpha & = \frac \pi 2 - \beta \\ \beta & = \frac \pi 2 - \alpha \end{aligned}$

Therefore, we have:

$\begin{aligned} f(x) & = \left(\sin^{-1}x\right)^3 + \left(\cos^{-1}x\right)^3 \\ & = \alpha^3 + \beta^3 \\ & = \alpha^3 + \left(\frac \pi 2 - \alpha \right)^3 \\ & = \alpha^3 + \frac {\pi^3}8 - \frac {3\pi^2}4 \alpha + \frac {3\pi}2 \alpha^2 - \alpha^3 \\ & = \frac {3\pi}2 \left( \alpha^2 - \frac \pi 2 \alpha + \frac {\pi^2}{12} \right) \\ & = \frac {3\pi}2 \left( \alpha^2 - \frac \pi 2 \alpha + \frac {\pi^2}{16} + \frac {\pi^2}{12} - \frac {\pi^2}{16} \right) \\ & = \frac {3\pi}2 \left( \left(\alpha - \frac {\pi}4\right)^2 + \frac {\pi^2}{48} \right) \\ & = \frac {3\pi}2 \left(\alpha - \frac {\pi}4\right)^2 + \frac {\pi^3}{32} & \small \color{#3D99F6} \text{Note that }\left(\alpha - \frac {\pi}4\right)^2 \ge 0 \\ \implies f_{min}(x) & = {\color{#3D99F6}0} + \frac {\pi^3}{32} & \small \color{#3D99F6} \text{when }\left(\alpha - \frac {\pi}4\right)^2 = 0 \text{, the minimum.} \\ \implies f_{max}(x) & = \frac {3\pi}2 \left({\color{#3D99F6} - \frac \pi 2} - \frac {\pi}4\right)^2 + \frac {\pi^3}{32} = \frac {7\pi^3}8& \small \color{#3D99F6} \text{when }\left(\alpha - \frac {\pi}4\right)^2\text{ is the maximum.} \end{aligned}$

Therefore $f_{min}(x) + f_{max}(x) = \dfrac {7\pi^3}8 + \dfrac {\pi^3}{32} = \dfrac {29\pi^3}{32}$ . $\implies a+b = 29+32 = \boxed{61}$ .

$let\quad \sin ^{ -1 }{ x=X\quad , } \cos ^{ -1 }{ x=Y } \quad \Rightarrow X+Y=\cfrac { \pi }{ 2 } \\ f\left( x \right) =X^{ 3 }+{ Y }^{ 3 }=\left( X+Y \right) \left[ { \left( X+Y \right) }^{ 2 }-3XY \right] \\ =\cfrac { \pi }{ 2 } \left[ \cfrac { { \pi }^{ 2 } }{ 4 } -3Y\left( \cfrac { \pi }{ 2 } -Y \right) \right] \\ =\cfrac { 3\pi }{ 2 } \left[ { \left( Y-\cfrac { \pi }{ 4 } \right) }^{ 2 }+\cfrac { { \pi }^{ 2 } }{ 48 } \right] \\ \therefore \quad for\quad Y\in \left[ 0,\pi \right] ,\\ { f }_{ max }={ f }_{ Y=\pi }=\cfrac { 7 }{ 8 } { \pi }^{ 3 }\\ \& \quad { f }_{ min }={ f }_{ Y=\cfrac { \pi }{ 4 } }=\cfrac { { \pi }^{ 3 } }{ 32 } \\ \Rightarrow { f }_{ max }+{ f }_{ min }=\cfrac { 29 }{ 32 } { \pi }^{ 3 }\quad \Rightarrow a+b=29+32=61$