How to find inverses of elements ( If you do not want to do computations)

I was reading Algebra:Ring theory and Field theory notes where I found the following exercise.

Suppose $\alpha \in \Bbb{C}$ is zero of $x^3-x+1$. Express $(\alpha)^{-1},(\alpha+1)^{-1}, (\alpha^2+1)^{-1}$ in terms of $a_0+a_1\alpha+a_2\alpha^2$ for $a_i\in\Bbb{Q}$.

Working : We know $Q[\alpha]\equiv Q[x]/\langle x^3-x+1\rangle$ which is a field.

So the above-mentioned should exist. We know $(\alpha)^{-1}$ can be expressed as $-\alpha^2+1$ and $(\alpha+1)^{-1}$ as $(-\alpha(\alpha-1))$. But I was not able to find $(\alpha^2+1)^{-1}$.

Turns out, it is very simple. We use extended euclidean algorithm!

So \(x^{3} - x + 1 = \bigg(x^{2} + 1\bigg)\bigg(x\bigg) + \bigg(-2 x + 1\bigg)\) \(x^{2} + 1 = \bigg(-2 x + 1\bigg)\bigg(-\frac{1}{2} x - \frac{1}{4}\bigg) + \bigg(\frac{5}{4}\bigg)\) \(-2 x + 1 = \bigg(\frac{5}{4}\bigg)\bigg(-\frac{8}{5} x + \frac{4}{5}\bigg) + \bigg(0\bigg)\)

\[\bigg(\frac{2}{5} x + \frac{1}{5}\bigg)\bigg(x^{3} - x + 1\bigg) + \bigg(-\frac{2}{5} x^{2} - \frac{1}{5} x + \frac{4}{5}\bigg)\bigg(x^{2} + 1\bigg) = 1\]

So inverse of $x^2+1$ is \(-\frac{2}{5} x^{2} - \frac{1}{5} x + \frac{4}{5}.\)

Alternatively, I liked this explanation in StackExchange!