<s>[INST] 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.
 [/INST] $$x=m-m^2-2=-(m+m^2+1)+2m-1=2m-1$$
And $2m-1 \neq1$, here is your mistake. Then we can continue our calculations :
$$f(x)=x^4+3x^3+2x^2-11x+6\\=16m^4-32m^3+24m^2-8m+1+24 m^3-36 m^2+18 m-3+8 m^2-8 m+2-22m+11+6\\=16 m^4-8 m^3-4 m^2-20 m+17\\=16m^2(m^2+m+1)-24m^3-20m^2-20m+17\\=-24m^3-20m^2-20m+17=-24m(m^2+m+1)+4m^2+4m+17\\=4m^2+4m+17\\=4(m^2+m+1)+13\\=13$$
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