Find inverse of element in a binary field The question states:
Let us consider the field $GF(2^4)$ with multiplication modulo $x^4+ x^3+1$
Find all y such that $1010(y + 0011) = 1111$, in other words find y that satisfies $(x^3+x)(y +x+1) = x^3+x^2+x+1$
I tried to find the inverse for element $1010$ to multiply both sides of the equation and continue from there, I tried by solving the equation:
$(x^3+x)(ax^3+bx^2+cx+d) = 1$
But the solution I get is that $a=0,b=1,c=1,d=0$ which doesn't produce the inverse element.
What is the inverse element of $(1010)$ in this field and how can I arrive at that solution?
Any help is greatly appreciated.

Here are some methods to compute this using a computer. The software is sagemath.
First, if the program supports defining finite fields with a given polynomial, you can just use that:
K.<a> = GF(2^4, modulus=x^4 + x^3 + 1)
(a^3 + a)^-1
  => a^3 + a + 1

You can also use the extended GCD which gives polynomials $u, v$ such that $uf + vg = 1$ so, modulo $f$, that gives $v = g^{-1}$.
K = GF(2)[x]
xgcd(K(x^4 + x^3 + 1), K(x^3 + x)) # K(...) means "as an element of K"
  => (1, x^2 + x + 1, x^3 + x + 1)

So $(x^3 + x + 1)(x^4 + x^3 + 1) + (x^3 + x + 1)(x^3 + x) = 1$.
You can also do it "by hand" using companion matrices. If $x^4 + ax^3 + bx^2 + cx + d$ is the minimal polynomial of your generator $\alpha$, then form the matrix
$$A = \begin{pmatrix}
0 & 0 & 0 & -d \\
1 & 0 & 0 & -c \\
0 & 1 & 0 & -b \\
0 & 0 & 1 & -a
\end{pmatrix}$$
In this way, $A$ also has that minimal polynomial. More specifically, each column represents $\alpha \cdot t$ for $t \in \{1, \alpha, \alpha^2, \alpha^3\}$. Now you can compute $(A^3 + A)^{-1}$, and again the columns represent $(\alpha^3 + \alpha)^{-1} \cdot t$ for $t$ in that ordered basis. Of course, you want $t = 1$ to find $(\alpha^3 + \alpha)^{-1}$.
Here $(a,b,c,d) = (1,0,0,1)$ and
$$(A^3 + A)^{-1} = \begin{pmatrix}
1&1&1&0\\
1&1&1&1\\
0&1&1&1\\
1&1&0&1
\end{pmatrix}.$$
Thus $(\alpha^3 + \alpha)^{-1} = 1 + \alpha + \alpha^3$ by reading the first column.
You can also just solve the linear system $(A^3 + A)u = (1,0,0,0)$ rather than the full inverse $(A^3 + A)U = I$. This works out to be the same linear system as given in GEdgar's answer.
