Find the inverse of $x^6+x^4+x+1$ in the finite field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$ 
How can I find the inverse of $x^6+x^4+x+1$ in the field $\mathbb{Z}_2[x]/(x^8+x^4+x^3+x+1)$?

can someone help me please to solve the problem.
actually I want to know the process of calculation.
I know that $1+(x^8+x^4+x^3+x+1)$ is the identity element and try to calculate another polynomial $a(x) $ such that $(x^6+x^4+x+1)a(x)=1$ but this is very lengthy process.
 A: You can use Extended Euclidean algorithm.
It only take three steps:
$x^8+x^4+x^3+x+1=(x^2+1)(x^6+x^4+x+1)+x^2$
$x^6+x^4+x+1=(x^4+x^2)x^2+x+1$
$x^2=(x+1)(x+1)+1$  
You also needs to express the remainders using $f(x)=x^8+x^4+x^3+x+1$ and $g(x)=x^6+x^4+x+1$.
Let us make this into a table.
Each line contains coefficients such that $a(x)f(x)+b(x)g(x)=r(x)$. The next line is always computed using the preceding two lines.
$$
\begin{array}{|c|c|c|c|c|}
\hline
   a(x) & b(x) & r(x) \\\hline
   1 & 0 & x^8+x^4+x^3+x+1 \\\hline
   0 & 1 & x^6+x^4+x+1 \\\hline
   1 & x^2+1 & x^2 \\\hline
   x^4+x^2 & x^6+x^2+1 & x+1 \\\hline
   x^5+x^4+x^3+x^2+1 & x^7+x^6+x^3+x & 1 \\\hline
\end{array}
$$
We have $(x^5+x^4+x^3+x^2+1)(x^8+x^4+x^3+x+1)+(x^7+x^6+x^3+x)(x^6+x^4+x+1)=1$ in $\mathbb Z_2[x]$.
A useful thing to do, when computing something like this, is to plug in some values into the polynomials (in this case we only have $x=0$ and $x=1$). There is at least some probability, that if you made a mistake, it might manifest already on the values you tried. (To be entirely sure, you would have to try to multiply the polynomials, which is rather cumbersome.)
You can find many other examples on this site for finding inverses in finite fields.
For example: Finding inverse of polynomial in a field.
Another post explaining table method is here: How to use the Extended Euclidean Algorithm manually? (There are certainly more such posts.)
