We have given one roots of the polynomial \(\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}-{4}\) is 1-i then other root will be 1+i since, roots of polynomial occur in complex conjugate.

Now, we have to find just one real roots.

For check x=2 we see

\(\displaystyle{x}^{{3}}-{4}{x}^{{2}}+{6}{x}={\left({2}\right)}^{{3}}-{4}{\left({2}\right)}^{{2}}+{6}{\left({2}\right)}-{4}\)

\(\displaystyle={8}-{16}+{12}-{4}\)

\(\displaystyle={0}\)

Therefore, x=2 is a real root of the function.

Therefore, another two roots of the polynomials are 2,1+i