$\mathbb{F}_{3}[x]/(x^3-x^2+1) \cong \mathbb{F}_{3}[x]/(x^3-x^2+x+1)$. I want to prove that $\mathbb{F}_{3}[x]/(x^3-x^2+1) \cong \mathbb{F}_{3}[x]/(x^3-x^2+x+1)$.
I have by Proposition 3.117 from Rotman´s Advanced Algebra that both are fields and we have that:
$$\mathbb{F}_{3}[x]/(x^3-x^2+1)= \lbrace a+ b\alpha / \alpha^3-\alpha^2+1=0 \rbrace$$
And;
$$\mathbb{F}_{3}[x]/(x^3-x^2+x+1)= \lbrace a+b \beta / \beta^3-\beta^2+\beta+1  \rbrace$$.
So how can I define a properly isomorphism between both fields? Thanks
 A: Think of $\mathbb{F}_{3}[x]/(x^3-x^2+1)$ as the algebra $A=\mathbb F_3[a]$ generated by $a$ satisfying the relation $a^3 - a^2+1=0$. And similarly, think of the second field as $B=\mathbb F_3[b]$ with $b$ satisfying $b^3-b^2+1=-b$. Now, to define a homomorphism from $A$ to $B$ it is necessary and sufficient to find an element in $B$ that satisfies the same relation over $\mathbb F_3$ as $a$. With $c:=b+1$, we have $c^3-c^2+1=b^3+1-b^2-2b-1+1=b^3-b^2+1-2b=-b-2b=-3b=0$ (where we have used the relation $b$ satisfies and the fact that we are working in characteristic $3$). Thus, the homomorphism sending $a$ to $c=b+1$ should do the trick. The inverse homomorphism should be obvious.
EDIT: Mike described the inverse of this homomorphism in his answer, his $\delta, \gamma$ correspond to my $a, b$ respectively. Note that $+2=-1$.
A: Both are 27-element fields. As all fields of the same size are isomorphic to each other, that established isomprhism.
As for defining this isomorphism explicitly, note the following: Let $\gamma$ be the root of $x^3-x^2+1$ in $\mathbb{F}_3$, and let
$\delta$ be a root of  $x^3-x^2+x+1$ in $\mathbb{F}_3$. Then $\mathbb{F}_3[x]/(x^3-x^2+1) = \mathbb{F}_3[\gamma]$ and $\mathbb{F}_3[x]/(x^3-x^2+x+1) = \mathbb{F}_3[\delta]$.
A mapping $\phi: \mathbb{F}_3[x]/(x^3-x^2+1) = \mathbb{F}_3[\gamma] \mapsto \mathbb{F}_3[x]/(x^3-x^2+x+1) = \mathbb{F}_3[\delta]$ that fixes $\mathbb{F}_3$ is an isomorphism iff the following holds:

Property 1: For each polynomial $p$ with coeeficients in $\mathbb{F}_3$, the following holds: $p(\gamma)=0$ iff $p(\phi(\gamma))=0$

However, note the following: both $\gamma$ and $\delta+1$ are roots of the polynomial $x^3-x^2+1$ in $\mathbb{F}_3$, and both $\delta$ and $\gamma+2$ are roots of the polynomial $x^3-x^2+x+1$ in $\mathbb{F}_3$. 
So letting $\phi(\gamma) = \delta+1$ [and $\phi(y)=y$ for each $y \in \mathbb{F}_3$], it follows that $\gamma$ and $\phi(\gamma)=\delta+1$ are both roots of the polynomial $x^3-x^2+1$ in $\mathbb{F}_3$. This is necessary and sufficient for  Property 1 to hold, as $x^3-x^2+1$ is irreducible.
Thus the mapping ; $\phi(\gamma) = \delta+1$ is the desired isomorphism.
A: Both $x^3-x^2+1$ and $x^3-x^2+x+1$ are irreducible over $\Bbb F_3$ (since  neither has a root).  Thus both quotients are the field $\Bbb F_{27}=\{\alpha+\beta x+\gamma x^2\mid \alpha,\beta,\gamma \in\Bbb F_3\}$.
