In Dummit and Foote's textbook one of the exercises is:
Let $\theta$ be a root of $x^3-2x-2$ over $\mathbb{Q}$. Compute $\frac{1+\theta}{1+\theta+\theta^2}$ in $\mathbb{Q}(\theta)$.
My approach was to look at $\theta^{-1}$: The polynomial gives: $$\theta^3-2\theta-2=0$$ $$2\theta^{-1}=\theta^2-2$$ Then I changed the expression to be simplified and substituted: $$\frac{1+\theta}{1+\theta+\theta^2}=\frac{2\theta^{-1}+2}{2\theta^{-1}+2+2\theta}$$ $$=\frac{\theta^2-2+2}{\theta^2-2+2+2\theta}$$ $$=\frac{\theta^2}{\theta^2+2\theta}$$ $$=\frac{\theta}{\theta+2}$$ And then I divided $\theta+2$ into $\theta^3-2\theta-2$ and scaled it to get: $$(\theta+2)^{-1}=\frac{1}{6}(\theta^2-2\theta+2)$$ And substituted that to get: $$\theta\frac{1}{6}(\theta^2-2\theta+2)$$ $$\frac{\theta^3-2\theta^2+2\theta}{6}$$ $$\frac{-\theta^2+2\theta+1}{3}$$
I have two questions: How do I know if this the right answer, and is there a less convoluted way to do this?
Because $x^3-3x-2$ is ireducible in $\mathbb{Q}[X]$ we have g.c.d of $x^3-3x-2$ with $1+x+x^2$ is $1$ i.e., there exists $f(x)$ and $g(x)$ such that $$f(x)(x^3-3x-2)+g(x)(x^2+x+1)=1$$
Then, See that $\theta^3-3\theta-2=0$ so, we have $g(\theta)=\frac{1}{\theta^2+\theta+1}$..
Then multiply by $1+\theta$ to get the required answer..