Let tthe polynominal \(\displaystyle{f{{\left({x}\right)}}}={x}+{1}{\quad\text{and}\quad}{p}{\left({x}\right)}={x}^{{3}}-{2}.\)

Using the Educlidean algorithm theorem, divide \(\displaystyle{x}^{{3}}-{2}\ {b}{y}\ {x}+{1}.\)

\(\displaystyle{x}^{{3}}-{2}={\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}-{3}\)

\(\displaystyle{x}^{{3}}-{2}-{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}=-{3}\)

Divide by -3 on the both the sides of the above equation.

\(\displaystyle{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{3}}-{2}\right)}-{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}={\left(-\frac{{3}}{{-{{3}}}}\right)}\)

\(\displaystyle{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{3}}-{2}\right)}+{\left(-\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({x}+{1}\right)}={1}\)

That is \(\displaystyle{\left\langle{p}{\left({x}\right)}\right\rangle}+{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\cdot{\left({f{{\left({x}\right)}}}\right)}={1}\)

Therefore, the inverse of the element \(x+1\) is \(\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}\)

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 \(\displaystyle{F}{\left(\alpha\right)}\stackrel{\sim}{=}{F}\frac{{{x}}}{{{p}{\left({x}\right)}}}'\)

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 \(\displaystyle{o}{f}{2}^{{\frac{{1}}{{3}}}}+{1}\) over \(\displaystyle\mathbb{R}\)

The inverse of the element \(x+1\) is \(\displaystyle{\left(\frac{{1}}{{3}}\right)}\cdot{\left({x}^{{2}}-{x}+{1}\right)}.\)