d
(d) \(2x = A({x^2} + x + 1)\, + (Bx + C)\,(x - 1)\)

For \(x = 1\), \(2 = 3A\) \( \Rightarrow \) \(A = {2 \over 3}\)

For \(x = \omega ,\,2\omega = A(1 + \omega + {\omega ^2}) + B{\omega ^2} + (C - B)\,\omega - C\)

\( \Rightarrow \) \(2\omega = A.0 + B{\omega ^2} + (C - B)\,\omega - C\)

\(\omega = {{ - 1 + \sqrt 3 i} \over 2},\,\,{\omega ^2} = {{ - 1 - \sqrt 3 i} \over 2}\)

\(\therefore  - 1 + \sqrt 3 i = B\left( { - {1 \over 2} - {{\sqrt 3 } \over 2}i} \right) + (C - B)\,\left( { - {1 \over 2} + {{\sqrt 3 } \over 2}i} \right) - C\)

\( \Rightarrow \)\( - 1 + \sqrt 3 i = \left( { - {B \over 2} - {C \over 2} + {B \over 2} - C} \right) + {{i\sqrt 3 } \over 2}(C - 2B)\)

\( \Rightarrow \)\( - 1 = - {3 \over 2}C,\,\sqrt 3 = {{\sqrt 3 } \over 2}(C - 2B)\)

\(C = {2 \over 3},\,B = {{C - 2} \over 2} = - {2 \over 3}\)

\(\therefore A = C \ne B\) \( \Rightarrow \) \(A \ne B \ne C\).