Given $F$ that is a subfield of $\mathbb{C}$, I want prove or disprove whether $f(x) + \langle x^3 -1 \rangle$ is a unit in the quotient ring $F[x]/ \langle x^3 -1 \rangle$, given (1) $f(x) = x^2 -1$ and (2) $f(x) = 3x^5$.
I understand that $F[x]/\langle x^3 - 1\rangle$ is not a field since $x^3 -1$ is reducible in any subfield of $\mathbb{C}$. So for both cases they might not have an inverse. I tried to work from definition by prove or disprove that there exists $g(x)$ such that $f(x)g(x) + \langle x^3 - 1\rangle = 1 + \langle x^3 - 1\rangle$, which is equivalent to $f(x)g(x) -1 = h(x)(x^3-1)$ for some $h(x)$. I got stuck here as I cannot prove or disprove the existence of $g(x)$ or $h(x)$ for both cases of $f(x)$. Any help would be appreciated.
The ring $F[x]/ \langle x^3 -1 \rangle$ consists of elements of the form $$\{a+bx+cx^2:a,b,c\in F, x^3=1\}$$
To find inverse of $3x^5+\langle x^3-1\rangle$, note that $$3x^5-3x^2=3x^2(x^3-1)\in\langle x^3-1\rangle$$, i.e. $3x^5+\langle x^3-1\rangle=3x^2+\langle x^3-1\rangle$ and its inverse is $\frac{x}{3}+\langle x^3-1\rangle$.
To find an inverse for $x^2-1+\langle x^3-1\rangle$, suppose the inverse is of the form $a+bx+cx^2+\langle x^3-1\rangle$, and we want to have $$(x^2-1+\langle x^3-1\rangle)(a+bx+cx^2+\langle x^3-1\rangle)=1+\langle x^3-1\rangle $$ or $$(x^2-1)(a+bx+cx^2)+\langle x^3-1\rangle=1+\langle x^3-1\rangle$$ But the product $(x^2-1)(a+bx+cx^2)$ is $$ax^2+bx^3+cx^4-a-bx-cx^2=ax^2+b+cx-a-bx-cx^2$$ because $x^3=1$ and $x^4=x(x^3)=x$ (Not that by $x^4=x$, we mean in the quotient ring, $x^4-x\in\langle x^3-1\rangle $, therefore $x^4+\langle x^3-1\rangle=x+\langle x^3-1\rangle$.)
Thus we need to have $$(b-a)+(c-b)x+(a-c)x^2=1$$ i.e. $b-a=1 $, $c=b$ and $a=c$ which has no solution, therefore $x^2-1+\langle x^3-1\rangle$ is not invertible.