Extended Euclidean Algorithm in $GF(2^8)$?

Using Rijndael's finite field, the reducing polynomial is $x^8+x^4+x^3+x+1$.

Suppose we want to compute the inverse of $x^5+1$ in this field. We want to solve the equation $$ a(x^5+1)+b(x^8+x^4+x^3+x+1)=1 $$ I like to use the Euclid-Wallis Algorithm. Since we are dealing with polynomials, I will write things rotated by $90^\circ$.

$$ \begin{array}{c|c} x^8+x^4+x^3+x+1&0&1&\\\hline x^5+1&1&0\\\hline x^4+x+1&x^3&1&x^3\\\hline x^2+x+1&x^4+1&x&x\\\hline \color{#C00000}{1}&\color{#C00000}{x^6+x^5+x^3+x^2+x}&\color{#C00000}{x^3+x^2+1}&x^2+x\\\hline 0&x^8+x^4+x^3+x+1&x^5+1&x^2+x+1 \end{array} $$ The fifth row tells us that in $\mathbb{Z}_2[x]$ $$ (x^5+1)(\color{#C00000}{x^6+x^5+x^3+x^2+x})+(x^8+x^4+x^3+x+1)(\color{#C00000}{x^3+x^2+1})=\color{#C00000}{1} $$ Thus, $x^6+x^5+x^3+x^2+x$ is the inverse of $x^5+1$ in $\left.\mathbb{Z}_2[x]\middle/(x^8+x^4+x^3+x+1)\mathbb{Z}_2[x]\right.$.