1. **State the problem:** Factor the polynomial $$x^{10} + x^5 + 1$$. 2. **Recognize the structure:** Notice that the polynomial resembles a quadratic in terms of $$x^5$$, since $$x^{10} = (x^5)^2$$. 3. **Set a substitution:** Let $$y = x^5$$, so the polynomial becomes $$y^2 + y + 1$$. 4. **Factor the quadratic:** The quadratic $$y^2 + y + 1$$ does not factor over the real numbers because its discriminant $$\Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0$$. 5. **Use complex roots:** The roots of $$y^2 + y + 1 = 0$$ are $$y = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$. 6. **Rewrite the factorization:** Over complex numbers, $$y^2 + y + 1 = (y - \omega)(y - \omega^2)$$ where $$\omega = e^{2\pi i/3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$$ and $$\omega^2$$ is its conjugate. 7. **Substitute back:** Replace $$y$$ with $$x^5$$ to get $$$ x^{10} + x^5 + 1 = (x^5 - \omega)(x^5 - \omega^2) $$$ 8. **Summary:** The polynomial factors over complex numbers as the product of two quintic polynomials with complex coefficients. **Final answer:** $$$ x^{10} + x^5 + 1 = (x^5 - e^{2\pi i/3})(x^5 - e^{4\pi i/3}) $$$