If $$x \neq y \neq z$$ with $$x + y + z = 0$$ and $$x^3 + x + 1 = 0$$, $$y^3 + y + 1 = 0$$, $$z^3 + z + 1 = 0$$, then the numerical value of $$x^9 + y^9 + z^9$$ is

Since $$x, y, z$$ are distinct and each satisfies $$t^3 + t + 1 = 0$$, they are the three roots of that cubic, so $$x + y + z = 0$$, $$xy + yz + zx = 1$$ and $$xyz = -1$$. Writing $$p_k = x^k + y^k + z^k$$, Newton's identities give $$p_1 = 0$$, $$p_2 = -2$$, $$p_3 = -3$$ and the recurrence $$p_k = -p_{k-2} - p_{k-3}$$ for $$k \geq 4$$. This yields $$p_4 = 2$$, $$p_5 = 5$$, $$p_6 = 1$$, $$p_7 = -7$$, $$p_8 = -6$$ and finally $$p_9 = -p_7 - p_6 = 7 - 1 = 6$$.