abstract algebra - How to find minimal polynomial in $GF(2^3)$

I have $GF(2^3)$ field defined by $\Pi(x)=x^3+x+1$. From literature they say those are the minimal polynomial, but I can't understand the operative method to find them. Any explanation for a general method?

$$\begin{array}{lll}
\textbf{Elem.} & \textbf{Polyn.} & \color{red}{\textbf{Minimal Polyn.}} \\
0 & 0 & \color{red}{x} \\
\alpha^0 & 1 & \color{red}{x+1} \\
\alpha^1 & \alpha & \color{red}{x^3+x+1} \\
\alpha^2 & \alpha^2 & \color{red}{x^3+x+1} \\

\alpha^3 & \alpha+1 & \color{red}{x^3+x^2+1} \\
\alpha^4 & \alpha^2+\alpha & \color{red}{x^3+x+1} \\
\alpha^5 & \alpha^2+\alpha+1 & \color{red}{x^3 + x^2 + 1} \\
\alpha^6 & \alpha^2+1 & \color{red}{x^3 + x^2 + 1} \\
\end{array}$$

Note: I saw another post about minimal polynomial but there was no such method explained

Answer

The elements of $GF(8)$ are exactly the roots of the polynomial $X^8-X\in GF(2)[X]$. This polynomial decomposes into irreducible polynomials as follows,
$$X^8-X = X(X+1)(X^3+X+1)(X^3+X^2+1).$$

If the field $GF(8)$ is given as $GF(2)[X]/\langle X^3+X+1\rangle$ and $\alpha$ is a zero of $X^3+X+1$, then (as stated above), $\alpha,\alpha^2,\alpha^4$ are the zeros of $X^3+X+1$ and $\alpha^3,(\alpha^3)^2=\alpha^6,((\alpha^3)^2)^2=\alpha^5$ are the zeros of $X^3+X^2+1$.