Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$ 
Invert $\overline{x+1}$ in $\mathbb Q[x]/(x^2+x+1)$.

So i know that  the coset representatives for $\mathbb Q[x]/(x^2+x+1) = \{a+bx : a,b \in \mathbb{Q}\}$. But I am unsure as to how to invert this. Any help? 
would $\overline{x+1} = \overline{(x^2+x+1) - x^2} = \overline{0} - \overline{x^2} = \overline{-x^2} $ work? I am not sure where to go from here.

We need to find some polynomial $P(x)$ so that $P(x)\cdot (x+1) = 1$.  In $\Bbb Q[x]/\langle x^2+x+1\rangle$ we have $x^2 = -x-1$, so there are no polynomials of degree 2 or higher.  The polynomial $P$ that we seek therefore has the form $q_1x+q_0$.  So we want $$(q_1x+q_0)(x+1) = 1\qquad \pmod{x^2+x+1}.$$
Expanding the left-hand side we get   $$q_1x^2+(q_1+q_0)x+ q_0 = 1$$
and then because $x^2 = -x-1$ we can reduce this to 
 $$\begin{align}q_1(-x-1)+(q_1+q_0)x+ q_0 &= \\
q_0x+ q_0-q_1 &= 1
 \end{align}$$
Equating coefficients on both sides, we find $q_0=0$ and $q_1 = -1$, so $$P(x) = -x.$$
To check, multiply $(x+1)\cdot -x = -x^2 - x = (x+1) - x = 1$  which is what we wanted.
