Compute the inverse of a given element in a field extension

Show that p(x) = x^3 + 9x + 6 is irreducible over \mathbb{Q}. Let \theta be a root of p(x). Find the inverse of 1+\theta in \mathbb{Q}(\theta).

p(x) is Eisenstein at 3, and so is irreducible, and thus \mathbb{Q}(\theta) \cong \mathbb{Q}[x]/(p(x)) is a field.

Note that every element of \mathbb{Q}(\theta) has the form a + b\theta + c\theta^2 for some rational numbers a,b,c. Suppose (1+\theta)(a+b\theta+c\theta^2) = 1; comparing coefficients, we see that a-6c = 1, a+b-9c = 0, and b+c = 0. Solving this system, we can see then that (1+\theta)^{-1} = \frac{1}{4}(\theta^2 - \theta + 10).

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