# what's the inverse of $1 + \alpha$, where $\alpha$ is a root of $p(x) = x^3 + 9x + 6$

Show that $p(x) = x^3 + 9x + 6$ is irreducible in $\mathbb Q[x]$. Let $\alpha$ be a root of $p(x)$. Find the inverse of $1 + \alpha$ in $\mathbb Q[x]$.

So as far as the irreducibility is conccerned we can use the Einseinstein criterion (p=3). But how can we find the inverse of $1 + \alpha$?

Thanks

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Presumably, you want the inverse of $1+\alpha$ in ${\bf Q}(\alpha)$, not in ${\bf Q}[x]$.
Divide $p(x)$ by $1+x$; $$p(x)=(1+x)q(x)+r$$ Then substitute $x=\alpha$, and maneuver a bit.
Hint $\$ Just as for integers, the extended Euclidean algorithm works, and terminates in one step since $\rm\:x+1\:$ has degree $1,\:$ i.e. $\rm\: a\:p + b\:(1\!+\!x) = 1\:$ $\rm\Rightarrow\: (1\!+\!x)^{-1}\!\equiv b\pmod p,\:$ i.e. in $\rm\:\Bbb Q[x]/(p).$