(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 $ $A \ne B \ne C$.