If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it says that answer is $1$. Please help me to identify my mistake.

Replacing $x=m-m^2-2$ in your given polynomial function, we have 
$$f(x)=m^8-4 m^7+11 m^6-19 m^5+24 m^4-21 m^3+23 m^2-15 m+28$$ as a result. Now, noticing that $m$ is a cube root of $-1$, this is equal to 
$$f(x)=m^2-4 m^1+11 m^0-19 m^2+24 m^1-21 m^0+23 m^2-15 m+28=$$
$$5m^2+5m+18=5(m^2+m+1)+13=\color{blue}{13}.$$
