So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got: $$1=(x+1)-1(x) =(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$ This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$ But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...
(EDIT) enter image description here This is an image of the EEA calculations
Judging from the calculation at the link you provided, you're taking $\ x\ $ to be a root of the polynomial $\ x^8 + x^4 + x^3 + x + 1\ $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $\ x^8 = x^4 + x^3 + x + 1\ $. As it happens, when I multiplied your putative inverse $\ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x\ $ by $\ x^6+1\ $ I didn't get $1$. I got $\ x^5\ $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $\ x^4 + x^3 +x^2 + x + 1\ $ rather than $\ x^4 + x^3 + x + 1\ $.