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
$x^{2007}+y^{2007}+z^{2007}$
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
$x^{2007}+y^{2007}+z^{2007}$
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 $3b^{669} = \dfrac{2^{669}}{3^{668}}$.
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
$x^{2007}+y^{2007}+z^{2007} = 3.$