Max - Min

Using quadratic formula and $\ p = 2\sqrt{2}\$ we get

$\begin{aligned} x &= \dfrac{4p \pm \sqrt{16p^2+16}}{8} \\ &= \dfrac{p \pm \sqrt{p^2 + 1}}{2} \\ &= \dfrac{2\sqrt{2} \pm 3}{2} \end{aligned}$ so $\alpha = \dfrac{2\sqrt{2} - 3}{2} \approx 0.0858 \qquad \beta = \dfrac{2\sqrt{2} + 3}{2} \approx 2.9142 \\ \\$

We derive $\ f(x) \$ to find its critical values:

$\begin{aligned} f'(x) = \dfrac{(x^2 + 1)(2) - (2x - p)(2x)}{(x^2 + 1)^2} &= 0 \\ 2x^2 + 2 - 4x^2 + 2xp &= 0 \\ 2x^2 - xp - 1 &= 0 \\ \\ x = \dfrac{p \pm \sqrt{p^2 + 4}}{2} = \dfrac{2\sqrt{2} \pm \sqrt{12}}{2} &= \sqrt{2} \pm \sqrt{3} \end{aligned}$

However, $\ \sqrt{2} + \sqrt{3} \approx 3.1463 \$ and $\ \sqrt{2} - \sqrt{3} \approx \text{-}0.3178$ , so neither is in our domain $[\alpha, \beta]$ , and our extrema will occur at the boundaries of our domain.

We compute

$f\left(\dfrac{2\sqrt{2} + 3}{2}\right) = \dfrac{4}{7 + 4\sqrt{2}} = \dfrac{4(7 - 4\sqrt{2})}{17} \qquad \text{and} \qquad f\left(\dfrac{2\sqrt{2} - 3}{2}\right) = \dfrac{\text{-}4}{7 - 4\sqrt{2}} = \dfrac{\text{-}4(7 + 4\sqrt{2})}{17}$

so

$g(2\sqrt{2}) = \dfrac{4(7 - 4\sqrt{2})}{17} - \dfrac{\text{-}4(7 + 4\sqrt{2})}{17} = \dfrac{56}{17} \implies a^3 + b^3 = 56^3 + 17^3 = \boxed{180529}$