Đặt \(\dfrac{{f\left( x \right) - 1}}{{{x^2} - x - 2}} = g\left( x \right) \Rightarrow f\left( x \right) = \left( {{x^2} - x - 2} \right)g\left( x \right) + 1\).

Khi đó ta có: \(\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{{x^2} - x - 2}} = 3 \Rightarrow \mathop {\lim }\limits_{x \to 2} g\left( x \right) = 3\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {{x^2} - x - 2} \right)g\left( x \right) + 1} \right] = 1\).

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{{f^3}\left( x \right) + 3f\left( x \right) - 4}}{{{x^2} - 2x}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left[ {f\left( x \right) - 1} \right]\left[ {{f^2}\left( x \right) + f\left( x \right) + 4} \right]}}{{x\left( {x - 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{x - 2}}.\mathop {\lim }\limits_{x \to 2} \dfrac{{{f^2}\left( x \right) + f\left( x \right) + 4}}{x}\end{array}\)

Theo bài ra ta có:

\(\begin{array}{l}\,\,\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{{x^2} - x - 2}} = 3\\ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{\left( {x - 2} \right)\left( {x + 1} \right)}} = 3\\ \Leftrightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{x - 2}}.\dfrac{1}{{x + 1}} = 3\\ \Leftrightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{x - 2}}.\mathop {\lim }\limits_{x \to 2} \dfrac{1}{{x + 1}} = 3\\ \Leftrightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{x - 2}}.\dfrac{1}{3} = 3\\ \Leftrightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 1}}{{x - 2}} = 9\end{array}\)

Vậy \(\mathop {\lim }\limits_{x \to 2} \dfrac{{{f^3}\left( x \right) + 3f\left( x \right) - 4}}{{{x^2} - 2x}} = 9.\dfrac{{1 + 1 + 4}}{2} = 27\).

Chọn B.