How do I find _all_ the roots of $f$ in $F_{27}$ hence show that $f$ splits in $F_{27}$ The following is a continuation of the linked question;
{Find $(1 + 2α)^{−1}$ in $F_{27}$.}
Let $α$ be a root of $1 + 2x + x^3 \in F_3[x]$.
I was asked to show the order of $α$ and find all of the roots of $f$ in $F_{27}$ and hence show that $f$ splits in $F_{27}$ and express $f$ as a product of linear factors in $F_{27}[X]$.
This polynomial is irreducible over $F_3$ as it has no linear factors.
Hence, $F_{27} = F_3[α]$.
I have solved for the ord($α$) as shown below and thus shown that $α$ is primitive.
The order of $α$ is a divisor of $27−1 = 26$. Thus, ord$(α)$ is $2, 13$ or $26$.
First, ord$(α)$ is not $2$; otherwise, $α$ would be $1$ or $−1$, neither of which is a root of $1 + 2x + x^3$.
Furthermore, we have $α^{13} = −1$ but does not equal $1$,
indeed $$α^{13} = α·α^3 ·(α^3)^3 = α·(−2α−1)·(−2α−1)^3 = α·(−2α−1)·(−8α^3 −1) = α·(−2α−1)·(α^3 −1) = α·(α−1)·(α−2) = α^3 + 2α = −1 $$
Thus, ord$(α) = 26$ and $α$ is a primitive element of $F_{27}$.
My problem
How do I find all the roots of $f$ in $F_{27}$ hence show that $f$ splits in $F_{27}$??
My answer must be in the form $λ_0 + λ_1α + λ_2α^2, λ_0,λ_1,λ_2 \in \mathbb Z_3$
I have found a solution computing the Galois Group which seems valid however it is a topic I have not yet covered so I would prefer an alternative method if possible.
Potential Partial Solution
The following could perhaps be used to show the splitting of $f$ in $F_{27}$
Let $n=\deg(f)$. If $f(x)$ is irreducible of degree $n$, then $f(x)\mid x^{p^{n}}-x$. $\mathbb{F}_{p^{n}}$ is the splitting field over $\mathbb{F}_{p}$  of $x^{p^{n}}-x$, a separable polynomial (use the fact $\mathbb{F}_{p^{n}}^{\star}$ is a group and use Lagrange's theorem to see that), $x^{p^{n}}-x=\prod\limits_{a \in \mathbb{F}_p^{n}}(x-a)$. Hence $f(x)$ splits into distinct linear factors in $\mathbb{F}_{p_{n}}$.
(marked in bold the part that I am unsure of)
All help is appreciated.

From a theoretical point of view, we know that there is one and only one extension of degree $3$ of $\mathbb{F}_3$ (up to isomorphism), so that any root of the polynomial must lie in that extension. 
And yes, that follows from Lagrange’s Theorem. We know that $\mathbb{F}_{27}^*$ is cyclic, since a finite multiplicative subgroup of a field must be cyclic. (It’s a group because every nonzero element has an inverse: it’s a field.) In fact, you have a generator. So every nonzero element of $\mathbb{F}_{27}$ satisfies $x^{26}-1=0$, and so every element of $\mathbb{F}_{27}$ satisfies $x^{27}-x=0$. Thus, $\mathbb{F}_{27}$ contains the splitting field of $x^{27}-x$, which being separable (since $f’ = -1\neq 0$) must have $27$ distinct roots. So $\mathbb{F}_{27}$ is in fact the splitting field of $x^{27}-x$, and so $x^{27}-x$ splits over $\mathbb{F}_{27}$. 
This argument in fact can be used to show that the field of order $p^n$ is unique up to isomorphism: it is the splitting field of $x^{p^n}-x$. 

Now, how do you express the other two roots in terms of $\alpha$? You can do some brute force computation by plugging in $\lambda_0 + \lambda_1\alpha + \lambda_2\alpha^2$ into $x^3+2x+1$, use the fact that $\lambda_i^3=\lambda_i$ and that $\alpha^3+2\alpha+1=0$, and find values of the $\lambda_i$ that solve the resulting equation; it will yield a linear system over $\mathbb{F}_3$ to be solved. This would be the straightforward but possibly work-intensive way of doing it. 
We can take a few ad hoc shortcuts. The characteristic is $3$ and every element of $\mathbb{F}_3$ satisfies $x^3=x$; so
$$(\lambda_0 + \lambda_1\alpha + \lambda_2\alpha^2)^3 = \lambda_0 + \lambda_1\alpha^3 + \lambda_2\alpha^6.$$
Now note that $\alpha^3 = -2\alpha-1 = \alpha-1$. We can quickly test elements of the form $x=\lambda_1\alpha+\lambda_0$ to see if we get lucky, and to find out to what conditions $\lambda_1$ and $\lambda_0$ would need to satisfy for $x^3$ to equal $x-1$.  This might be worth doing because as soon as we find two of the roots, we can probably solve for the third even if it is not of this form. So if we get lucky and we find some other root besides $\alpha$, that will do.
We have:
$$\begin{align*}
\lambda_1\alpha + \lambda_0-1 &= (\lambda_1\alpha + \lambda_0)^3\\
&= \lambda_1\alpha^3 + \lambda_0\\
&= \lambda_1(\alpha - 1) + \lambda_0\\
&= \lambda_1\alpha + (\lambda_0-\lambda_1).
\end{align*}$$
This yields that we must have $\lambda_1=1$, but $\lambda_0$ arbitrary. That seems to suggest that $\alpha$, $\alpha+1$, and $\alpha+2$ will be the three roots. 
Indeed,
$$(\alpha+\lambda_0)^3 - (\alpha+\lambda_0) + 1 = \alpha^3+\lambda_0 - \alpha - \lambda_0 + 1 = \alpha^3 -\alpha + 1= 0.$$
So that gives you the three roots.
Or,
$$\begin{align*}
(x-\alpha)(x-\alpha+1)(x-\alpha-1)&= (x-\alpha)((x-\alpha)^2-1)\\
&= (x-\alpha)^3 - (x-\alpha)\\
&= x^3 - \alpha^3 - x + \alpha\\
&= x^3 - (\alpha-1) - x + \alpha\\
&= x^3 -x + 1\\
&= x^3 + 2x + 1.
\end{align*}$$
