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I'm trying to determine the minimal polynomials of $-\alpha$, $1-\alpha$, $2\alpha$, and $1/\alpha$, given that the minimal polynomial of $\alpha$ is $x^3-x-1$.

Since $x^3-x-1$ is the minimal polynomial of $\alpha$, we have $\alpha^3-\alpha-1=0$. So, $\alpha^3=\alpha+1$, and $-(\alpha^3-\alpha-1)=0$, so $-\alpha^3+\alpha+1=0$, and thus $-\alpha$ satisfies $x^3+x+1=0$. Would that be the minimal polynomial? How would I get the other minimal polynomials requested? Thanks!

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Hint $\ $ Change variables: $\ \beta = a\alpha+b\,\Rightarrow\ \alpha = (\beta-b)/a,\,$ so $\, 0 = f(\alpha) = f((\beta-b)/a) = g(\beta)\,$ and $\,f\,$ irreducble $\,\Rightarrow\, g\,$ irreducible.

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you can solve in this way: let $\beta=f(\alpha)$ write $\alpha=g(\beta)$ and put g into the minimal polinomial, the result would be a function of $\beta$ if it is not a polinomial then transform it, this is the minimal polinomial you look for. For example $\beta=1-\alpha$ so $\alpha=1-\beta$ and $(1-\beta)^3-(1-\beta)-1=0$ you just have to solve it.

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