Find the inverse of x+1 in QQ[x]/(x^3-2). Explain why this is the same as finding the inverse of root(3)(2) in RR.

Let tthe polynominal ${\displaystyle f\left(x\right)=x+1\text{and}p\left(x\right)=x^3-2.}$
Using the Educlidean algorithm theorem, divide ${\displaystyle x^3-2byx+1.}$
${\displaystyle x^3-2=(x^2-x+1)\cdot (x+1)-3}$
${\displaystyle x^3-2-(x^2-x+1)\cdot (x+1)=-3}$
Divide by -3 on the both the sides of the above equation.
${\displaystyle (-\frac{1}{3})\cdot (x^3-2)-(-\frac{1}{3})\cdot (x^2-x+1)\cdot (x+1)=(-\frac{3}{-3})}$
${\displaystyle (-\frac{1}{3})\cdot (x^3-2)+(-\frac{1}{3})\cdot (x^2-x+1)\cdot (x+1)=1}$
That is ${\displaystyle ⟨p\left(x\right)⟩+\left(\frac{1}{3}\right)\cdot (x^2-x+1)\cdot \left(f\left(x\right)\right)=1}$
Therefore, the inverse of the element $x+1$ is ${\displaystyle \left(\frac{1}{3}\right)\cdot (x^2-x+1)}$
By the theorem, "Let F be a field and let ${\displaystyle p\left(x\right)\in F\left(x\right)}$ be an irreducible polynominal. Suppose K is an extension field of F containing the root ${\displaystyle \alpha }$ of p(x), ${\displaystyle p\left(\alpha \right)=0}$. Let ${\displaystyle F\left(\alpha \right)}$ denote the subfield of K generated by ${\displaystyle \alpha }$ over F. Then ${\displaystyle F\left(\alpha \right)\stackrel{\sim }{=}F{\frac{x}{p\left(x\right)}}^{\prime }}$
Here $F=Q$ and ${\displaystyle p\left(x\right)=x^3-2}$
Note that, the root of the polynominal p(x) is ${\displaystyle {\left(2\right)}^{\frac{1}{3}}}$ and p(x) is ireeducible over Q
By the theorem ${\displaystyle Q\left({\left(2\right)}^{\frac{1}{3}}\right)\stackrel{\sim }{=}Q\frac{x}{x^3-2}}$
Therefore, the inverse of $x+1$ over ${\displaystyle Q\frac{x}{x^3-2}}$ sameas inverse ${\displaystyle of{2}^{\frac{1}{3}}+1}$ over ${\displaystyle \mathbb{R}}$
The inverse of the element $x+1$ is ${\displaystyle \left(\frac{1}{3}\right)\cdot (x^2-x+1).}$