<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] If $m$ is a primitive cube root of unity, $m^2+m+1=0$ is correct, but that just gives
$$ x = m-m^2-2 = -2\pm i\sqrt{3}. $$
Given $p(z)=z^4+3z^3+2z^2-11z+6$, we have:
$$ q(z) = p(z-2) = z^4-5z^3+8z^2-15z+28 $$
and
$$ p(x) = q(\pm i\sqrt{3}) =\color{red}{13}.$$
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