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?
 A: A general element of this ring looks like $a + bx + cx^2$. Suppose this were the inverse of $1+x^2$. Then:
\begin{align*}
1 &= (1+x^2)(a+bx+cx^2)\\
&= a + bx + (a+c)x^2 + bx^3 +cx^4
\end{align*}
In this ring, $x^3 = 5$ so:
\begin{align*}
1 &= a + bx + (a+c)x^2 +5b + 5cx \\
&=(a+5b) + (b+5c)x + (a+c)x^2.
\end{align*}
This tells us that $$a + 5b =1\\ b+5c = 0 \\ a + c = 0$$
I trust you can take it from here.

I can't resist adding an alternative approach. If you don't have a background in linear algebra, or this seems arcane, feel free to disregard it. 
Let's consider your ring as a vector space over $\mathbb{Q}$. Then it is three dimensional, and an easy basis would be $1,x,x^2$. Define the linear transformation on this vector space that multiplies a polyomial by $x$. Now let us find the matrix (call it $A$) of this transformation with respect to the basis above:
$$A = \begin{pmatrix} 0 & 0 & 5 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}_.$$
The cool thing is that if you want to find the inverse of $x^2+1$, you can instead find the inverse of the matrix $A^2 + I$ and then express it as a linear combination of $I, A, A^2$! In this case we have that:
\begin{align*}
(A^2 + I)^{-1} &= \frac{1}{26}\begin{pmatrix} 1 & -5 & 25 \\ 5 & 1 & -5 \\ -1 & 5 & 1 \end{pmatrix} \\
&= \frac{1}{26} I + \frac{5}{26} A - A^2.
\end{align*}
This tells us the the inverse of $x^2+1$ is $$\frac{1}{26} + \frac{5}{26}x - \frac{1}{26}x^2.$$
A: Use the Euclidean algorithm,

$$p(x)(x^2+1) + q(x)(x^3-5)=1$$

$$x^3-5 = x(x^2+1)+(-x-5)$$
$$x^2+1 = -x(-x-5) +5(-x-5)+26$$
$$-x-5 = -{x\over 26}(26)-4$$
$$26 = {2\over 13}(4)+0$$
Now as usual reverse the process to produce a polynomial inverse modulo $x^3-5$.
A: 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 $\alpha^{-1} = \dfrac{1}{26}(1 + 5x - x^2) + \langle x^3 - 5\rangle$.
