Least value -2

This question can be easily solved by observing that we can take $t=\sqrt{4-{x}^{2}}$ where $0 \le t \le 2$ and our function becomes :

$f(x)=g(t)={(3-t)}^{2}+{(1+t)}^{3}={t}^{3}+4{t}^{2}-3t+10$

Now we will check the value of the function at the extreme points as well as the critical points :

For finding critical points we will differentiate the function with respect to $t$ .

$\frac{dg(t)}{dt}=3{t}^{2}+8{t}-3=0$

$\Rightarrow t=\frac{1}{3}$

( $t$ can't be $-3$ )

We have $g(0)=10 , g(2)=28 , g(\frac{1}{3})=\frac{256}{27}$

Hence we conclude $min(f(x))=min(g(t))=\frac{256}{27}$

$\begin{aligned} f(x) & = \left(3-\sqrt{4-x^2}\right)^2 + \left(\color{#3D99F6}1+\sqrt{4-x^2}\right)^3 & \small \color{#3D99F6} \text{Let }u = 1+\sqrt{4-x^2} \\ g(u) & = (4-{\color{#3D99F6}u})^2 + {\color{#3D99F6}u}^3 \\ g'(u) & = 2(4-u)(-1)+3u^2 \\ & = 3u^2+2u-8 \\ g''(x) & = 6u + 2 \end{aligned}$

Equating $g'(u) = 0$ , we have:

$\begin{aligned} 3u^2+2u-8 & = 0 \\ (3u-4)(u+2) & = 0 \end{aligned}$

We note that $\begin{cases} g'' \left(\dfrac 43\right) = 6 \times \dfrac 43 + 2 = 10 > 0 & \implies g \left(\dfrac 43\right) \text{ is minimum.} \\ g''(-2) = 6(-2) + 2 = -10 < 0 & \implies g(-2) \text{ is maximum.} \end{cases}$

Therefore, $\min f(x) = \min g(u) = g \left(\dfrac 43\right) = \left(4-\dfrac 43\right)^2 + \left(\dfrac 43\right)^3 = \left(\dfrac 83\right)^2 + \left(\dfrac 43\right)^3 = \dfrac {256}{27}$ $\implies a+b = 256 + 27 = \boxed{283}$