So the idea is that arithmetic in $GF(2)$ is the same as arithmetic in the quotient of $(\mathbb Z/2\mathbb Z)[x]$ by $(x^8+x^7+x^3+x+1)$.
Remember to compute the multiplicative inverse of a number, say $7$ in the field $\mathbb Z$ quotiented by $11\mathbb Z$, we can solve the diophantine equation $7a + 11b = 1$ with $(a,b)$ in $\mathbb Z$ using EEA. It's the same here, we find the multiplicative inverse by solving the equation $(x^7+x^2+1)a + (x^8+x^7+x^3+x+1)b = 1$ with $(a,b)$ in $(\mathbb Z/2\mathbb Z)[x]$.
The best way to use EEA in practice (for numbers as well as polynomials) is by BlankinShip's Algorithm. I like that idea of writing the polynomials as 10000101
and 110001011
so let's use that notation.
/ 10000101 1 0 \ \ 110001011 0 1 / the first row op is to subtract the second row by 11 times the first / 10000101 1 0 \ \ 100 11 1 / now we need to subtract the first row by 100001 times the first / 1 1100010 100001 \ \ 100 11 1 / and we are done! So let us just check (if we were confident in our ability it would not be necessary to check) that 10000101*1100010 + 110001011*100001 = 1 it does, so now reducing this equation mod 110001011 gives us
$(x^7+x^2+1)\cdot(x^6+x^5+x) = 1 $ in GF(2)
notes on programming: If you implement polynomials in this binary way then addition would be implemented by the XOR
operation. Overflow should not be a problem because the numbers only decrease in size during execution of the algorithm.