Algebraic over $\Bbb Q$

You don't need to do any polynomial division. In part (i), you have proved $\Bbb{Q}(\alpha)$ has $1, \alpha, \alpha^2$ and $\alpha^3$ as a basis over $\Bbb{Q}$. In part (ii) you use this: to invert $1 + \alpha^3$, you have to solve: $$ \begin{align*} 1 &= (1 + \alpha^3)(a_0 + a_1\alpha + a_2\alpha^2 + a_3\alpha^3)\\ &= (a_0 + 20a_1) + (a_1 + 20a_2)\alpha + (a_2 + 20a_3)\alpha^2 + (a_3 + a_0)\alpha^3 \end{align*} $$ where the second equation follows by multiplying out and using $\alpha^4 = 20$. By part (i), the coefficients of $\alpha^3$, $\alpha^2$ and $\alpha$ must be zero and the constant term must be $1$, so you must have: $$ \begin{array}{rllr} a_3 &= -a_0 &\\ a_2 &= -20a_3 &= 20a_0\\ a_1 &= -20a_2 &= -400a_0\\ a_0 &= 1 -20a_1 &= 1 + 8000a_0 \end{array} $$ Now you can easily solve the last equation for $a_0$ and read off the other $a_i$, to get $$ (1 + \alpha^3)^{-1} = \frac{1}{7999}(-1 + 400\alpha -20\alpha^2 + \alpha^3). $$

Computing $(1+\alpha^3)^{-1}$ looks a little hairy using the extended Euclidean algorithm, but you can pull a little trick: first compute $(\alpha+20)^{-1}$ since that will terminate in a single long division, and then note that $\alpha(\alpha+20)^{-1}=(1+\alpha^3)^{-1}$

So, to be explicit:

$\alpha^4 - 20 = (\alpha^3 - 20 \alpha^2 + 400 \alpha - 8000) (\alpha + 20) + 159980$,

and you already have

$(\alpha + 20)^{-1}\equiv \frac{\alpha^3 - 20 \alpha^2 + 400 \alpha - 8000}{-159980}$

Then multiplying by $\alpha$ you get

$$ (1+\alpha^3)^{-1}\equiv\frac{\alpha^4 - 20 \alpha^3 + 400 \alpha^2 - 8000\alpha}{-159980}\\ \equiv\frac{20- 20 \alpha^3 + 400 \alpha^2 - 8000\alpha}{-159980}\\ \equiv\frac{1- \alpha^3 + 20 \alpha^2 - 400\alpha}{-7999} $$