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