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.

We know $$x^4+x^3+1=0 .\tag1$$
It will be useful to know
$$
x^4 = x^3+1\tag2$$
$$
x^5 = x^4+x=(x^3+1)+x = x^3+x+1\tag3$$
$$
x^6 = x^5+x^2 = (x^3+x+1)+x^2 = x^3+x^2+x+1\tag4
$$
These will be useful for doing a calculation like
$(x^3+x)(ax^3+bx^2+cx+d) = 1$ that the OP proposes.
Compute
\begin{align}
1 &= (x^3+x)(ax^3+bx^2+cx+d) 
\\
&=(a)x^6+(b)x^5+(c+a)x^4+(d+b)x^3+(c)x^2+(d)x+0
\\
&=(a+b+d+c+a+b)x^3+(a+c)x^2+(a+b+d)x+(a+b+c+a)
\\
&=(d+c)x^3+(a+c)x^2+(a+b+d)x+(b+c)
\end{align}
Thus
\begin{align}
d+c&=0\\a+c&=0\\a+b+d&=0\\b+c&=1
\end{align}
So $d=c$, then $a=c$, then $b=a+d=c+c=0$, then $c=d=a=1$.  So
$$
(x^3+x)^{-1} = x^3+x+1
$$
