Let $f(x)=x^3-x+1$ over $\mathbb{F_3}$
$\mathbb{F_3}$ is the finite field of order $3$, which is isomorphic to $\mathbb{Z_3}$.
$f(x)$ has no roots in $\mathbb{Z_3}$
Since it is a degree 3 polynomial, showing it has no roots in $\mathbb{Z_3}$ is enough to show it is irreducible over $\mathbb{Z_3}$
So the splitting field is $\dfrac{\mathbb{F_3}[x]}{\langle f(x) \rangle}$, where $\langle f(x) \rangle$ is the ideal generated by $f(x)$, which is a maximal ideal since $f(x)$ is irreducible, making this a field.
Now, setting $f(x)=0$, we get $x^3=x-1$ so $x^3$ and its higher powers can be replaced by $1,x,x^2$ in this field. So elements of this field are of the form $ax^2+bx+c+\langle f(x) \rangle$ where $a,b,c \in \mathbb{Z_3}$. So number of elements in this field is $3^3=27$, which is the required degree.
Is this correct?