Complex Numbers $x,y,z$ Find $x^{2007}+y^{2007}+z^{2007}$ Let $x,y,z$ be complex numbers such that
$$x+y+z = x^{5}+y^{5}+z^{5} = 0, \hspace{10pt}

x^3+y^3+z^3=2$$
Find all possible values of 
$$x^{2007}+y^{2007}+z^{2007}$$
 A: It $x+y+z$ = 0, then $x,y,z$ are roots of $t^3 + at-b = 0$.
Then, we can show that, $x^5 + y^5 + z^5 = -5ab$ and $x^3 + y^3 + z^3 = 3b$, either using Newton's Identities or as in my answer here: https://math.stackexchange.com/a/115534/1102
Since $b \ne 0$, we must have that $a = 0$.
Thus $x,y,z$ are roots of $t^3 = b$. We know that $b = \frac{2}{3}$ and thus can compute the expression you need easily as $3b^{669} = \dfrac{2^{669}}{3^{668}}$.
A: Since $x+y+z=0$,
$$x+y = -z\tag{A}.$$  
If we raise to power $3$ both sides
$$ x^3+3x^{2}y+3y^{2}x+y^{3} = -z^{3} \quad \Rightarrow \quad x^{3}+y^{3}+z^{3} = -3xy(x+y).$$
Since $x^{3}+y^{3}+z^{3} =3$, 
$$ x^{3}+y^{3}+z^{3} = -3xy(x+y).$$
We can conclude therefore that 
$$ xy(x+y) = -1 \tag{B}.$$
If we take fifth power to $(A)$,
$$ x^5+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5} = -z^{5} $$
$$ x^{5}+y^{5}+z^{5} = -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}).$$
And since $x^{5}+y^{5}+z^{5}=0$
$$ -5xy(x^{3}+2x^{2}y+2xy^{2}+y^{3}) = 0.  $$
Since $xy \neq 0$, 
$$ 
\begin{align*}
x^{3}+2x^{2}y+2xy^{2}+y^{3} &= 0\\
x^{3}+y^{3}+2xy(x+y) &= 0.
\end{align*}
$$
From $(B)$, 
$$ x^{3}+y^{3} = 2.$$  Also 
$$x^{3}+y^{3}+z^{3} = 1 \quad \Rightarrow \quad z^{3} = 1.$$
By symmetry
$$x^{3}=y^{3}=z^{3} = 1.$$
Therefore
$$x^{2007}+y^{2007}+z^{2007} = 3.$$
