Inverse element of 
Given the extension $ \dfrac{\mathbb Q[x]}{\langle x^3-5\rangle}: \mathbb Q$ and $\alpha =1+x^2+\langle x^3-5\rangle \in \dfrac{\mathbb Q[x]}{\langle x^3-5 \rangle}$, find $\alpha^{-1}$.

Can somebody give a hint? or an answer? please?

Since $x^3-5$ is a polynomial of degree $3$, every element in $k:= \dfrac{\mathbb{Q}[x]}{\langle x^3-5\rangle}$ has a representative of degree at most $2$. Thus we are looking for a polynomial $\beta:=a + bx + cx^2 + \langle x^3 - 5\rangle$ where $a,b,c\in \mathbb{Q}$ and such that $\alpha\beta = 1 + \langle x^3 + 5\rangle$.
Now
\begin{align*}\alpha\beta &= a + bx + (a+c)x^2 + bx^3 + cx^4 + \langle x^3 -5\rangle \\ &= (a+5b) + (b+5c)x + (a+c)x^2 + \langle x^3 -5\rangle
\end{align*}
Therefore
$$
\begin{cases}
a+5b = 1 & \\ b + 5c = 0 & \\ a + c = 0.
\end{cases}
$$
Solving this we get
$$a = \dfrac{1}{26} \qquad b = \dfrac{5}{26} \qquad c = \dfrac{-1}{26}$$
So $\alpha^{-1} = \dfrac{1}{26}(1 + 5x - x^2) + \langle x^3 - 5\rangle$.
