Multiplicative inverse in $GF(8)$

Question

I've the following problem:

$GF(8)$ is the Galois Field defined by $G(x) = x^3+x+1$.
$a(x) \in GF(8)$ is $a(x) = x+1$. What is the multiplicative inverse of $a(x)$?

I think it should be applied the Extended Euclidian Algorithm but I do not how to do it since the grade of $a(x)$ is lower the the $G(x)$ one.

Thank you all and sorry for the trivial question!

Answer 1

Not using brute force one can try the following: $GF(8)\simeq \Bbb{F}_2[x]/\langle x^3+x+1\rangle$. We use euclidean extended algorithm. Since $\operatorname{gcd}(x^3+x+1,x+1)=1$, we shall find such $a(x),b(x)$ for which $a(x)(x^3+x+1)+b(x)(x+1)=1$, meaning $b(x)(x+1)=1$ in $GF(8)$ (where all calculations are done in $\Bbb{F}_2[x]$).

So, $x^3+x+1=(x^2+x)(x+1)+1$ and therefore $x^3+x+1-(x^2+x)(x+1)=1$. $-(x^2+x)=x^2+x$ (in $F_2[x]$) so our inverse is $x^2+x$.