<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] As @David said, $x^2=4ω^2-4ω+1=-8ω-3=-4x-7$
Note that  $$x+1=2ω \Leftrightarrow x^4+3x^3+3x^2+x=8x \Leftrightarrow  x^4+3x^3=7x-3x^2$$
This implies that $$x^4+3x^3+2x^2-11x-6=-x^2-4x-6=1$$ 
However, another way to appraoch it using @David's method is by noting that $$x^4+3x^3+2x^2-11x-7+1=(x^2+4x+7)(x^2-x-1)+1=1$$. 
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