Not a Cube Root

First note that $x^{10} = x^{9} \times x = x$ , so the given expression simplifies to

$x^{8} + x^{5} + x^{2} = x^{2}(x^{6} + x^{3} + 1)$ .

But $x^{9} - 1 = (x^{3} - 1)(x^{6} + x^{3} + 1)$ , and since $x^{3} \ne 1$ we have that

$x^{6} + x^{3} + 1 = \dfrac{x^{9} - 1}{x^{3} - 1} = 0$ since $x^{9} = 1$ .

This in turn implies that the given expression equals $x^{2} \times 0 = \boxed{0}$ .

We can also go the complex number route and say that $x = e^{i \frac{2 \pi}{9}}$ is a solution. Then, considering the real and imaginary parts, we have:

$Re(e^{i \frac{20 \pi}{9}} + e^{i \frac{16 \pi}{9}} + e^{i \frac{10 \pi}{9}} + e^{i \frac{4 \pi}{9}} - e^{i \frac{2 \pi}{9}}) = 0$

$Im(e^{i \frac{20 \pi}{9}} + e^{i \frac{16 \pi}{9}} + e^{i \frac{10 \pi}{9}} + e^{i \frac{4 \pi}{9}} - e^{i \frac{2 \pi}{9}}) = 0$

which translates into

$cos(\frac{20 \pi}{9}) + cos(\frac{16 \pi}{9}) + cos(\frac{10 \pi}{9}) + cos(\frac{4 \pi}{9}) - cos(\frac{2 \pi}{9}) = 0$

$i(sin(\frac{20 \pi}{9}) + sin(\frac{16 \pi}{9}) + sin(\frac{10 \pi}{9}) + sin(\frac{4 \pi}{9}) - sin(\frac{2 \pi}{9})) = 0$