Inverse of $1-g-g^{-1}$ in group ring $\mathbb{Z}[G]$ for $o(g)=5$.

Question

I'm trying to prove that if $G$ is a group and $g\in G$ is an element of order $5$ then $x=1-g-g^{-1}$ is an unit element in the group ring $\mathbb{Z}[G]$. I'm trying to find an explicit inverse by computing $xg^r$ for $r=1,2,3,4$ and trying to add them in a good way to get $1$. But so far I can't find a good "combination". Any hints on how to compute an inverse for this element?.

[EDIT]

I found a solution for this but it is probably the hardest way. I'm trying to find an inverse of the form $a+bg+cg^2+dg^3+eg^4$, so i take the equation $x(a+bg+cg^2+dg^3+eg^4)=1$ and it becomes a linear algebra problem. I found by solving the $5\times 5$ integer linear system that $a=1,b=-1,c=0,d=-1,e=0$ is a solution, i.e., $x(1-g-g^4)=1$. I think there should be other ways to approach these kind of exercises.

Answer 1

Consider $x=g+g^{-1}$ and $y=g^2+g^{-2}$: note that $$ xy=g^3+g^{-1}+g+g^{-3}=g^{-2}+g^{-1}+g+g^2=x+y $$ Thus $$ (1-x)(1-y)=1-x-y+xy=1 $$

Answer 2

We have $x=1-g-g^4$. The extended Euclidean algorithm for $\gcd(1-g-g^4,g^5-1)$ gives $$ 1 = (1 - g^2 - g^3)(1-g-g^4)+(-g - g^2)(g^5-1) $$ Therefore, the inverse of $x$ is $1 - g^2 - g^3$ because $g^5-1=0$.