<s>[INST] If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is? 
If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is? ($\omega$ represents the cube roots of unity not equal to $1$).

Directly substituting the given value will work. But there is no fun in that. The calculations will also get lengthy.
$$\omega-\omega^2-2=2\omega-1$$
I tried simplifying the given equation.
$$x^4+3x^3+2x^2-11x-6=x^3(x+3)+2\left(x+\frac{1}{2}\right)(x-6)$$ 
I got stuck here.
 [/INST] Another approach might be to note that  $ \ \omega^2 \ = \ \overline{\omega} \ $ , thus  $ \ \omega \ - \ \omega^2 \ = \ 2 \ Im(\omega) \ = \  i \ \sqrt{3} \ $ .  So $ \ x \ + \ 2 \ = \ i \ \sqrt{3} \ $ .
The polynomial can be decomposed as
$$x^4 \ + \ 3x^3 \ + \ 2x^2 \ - \ 11x \ - \ 6 \ $$
$$ = \ (x \ + \ 2)^4 \ - \ 5(x \ + \ 2)^3 \ + \ 8(x \ + \ 2)^2 \ - \ 15(x \ + \ 2) \ +  \ 16  \ \ , $$
so its value for our choice of $ \ x \ $  is
$$  ( i \ \sqrt{3})^4 \ - \ 5( i \ \sqrt{3})^3 \ + \ 8( i \ \sqrt{3})^2 \ - \ 15( i \ \sqrt{3}) \ + \ 16 $$
$$  =  3^2 \ - \  [ (-i) \ 15 \sqrt{3}]  \ + \ 8( - 3) \ - \ ( i \ 15 \sqrt{3}) \ + \ 16 $$
$$  =  9 \ -24 \  + \ 16 \ = \ 1 \ \ . $$
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