1. **State the problem:** Solve the quadratic equation $$x^2 + x + 1 = 0$$ and analyze the expressions $$\frac{5}{x} + \frac{1}{x^5}$$. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Apply the formula:** Here, $$a=1$$, $$b=1$$, and $$c=1$$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3$$. Since $$\Delta < 0$$, the roots are complex: $$x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$. 4. **Express the roots:** $$x_1 = \frac{-1 + i\sqrt{3}}{2}, \quad x_2 = \frac{-1 - i\sqrt{3}}{2}$$. 5. **Evaluate $$\frac{5}{x} + \frac{1}{x^5}$$:** Since $$x$$ satisfies $$x^2 + x + 1 = 0$$, we can use this to simplify powers of $$x$$. From the equation: $$x^2 = -x - 1$$. Multiply both sides by $$x^3$$: $$x^5 = x^3 \times x^2 = x^3(-x - 1) = -x^4 - x^3$$. We need expressions for $$x^3$$ and $$x^4$$: Using $$x^2 = -x - 1$$: $$x^3 = x \times x^2 = x(-x - 1) = -x^2 - x = -(-x - 1) - x = x + 1 - x = 1$$. $$x^4 = x \times x^3 = x \times 1 = x$$. Therefore: $$x^5 = -x^4 - x^3 = -x - 1$$. 6. **Calculate $$\frac{1}{x^5}$$:** $$\frac{1}{x^5} = \frac{1}{-x - 1} = -\frac{1}{x + 1}$$. 7. **Calculate $$\frac{5}{x} + \frac{1}{x^5}$$:** $$\frac{5}{x} + \frac{1}{x^5} = \frac{5}{x} - \frac{1}{x + 1}$$. 8. **Simplify further:** Using the original equation, $$x^2 + x + 1 = 0$$, rearranged as $$x^2 = -x - 1$$. Multiply both sides by $$x$$: $$x^3 = 1$$ (as shown above). Since $$x^3 = 1$$, $$x$$ is a cube root of unity (not equal to 1). Recall the identity for cube roots of unity: $$1 + x + x^2 = 0$$. Therefore, $$x + 1 = -x^2$$. So: $$\frac{1}{x + 1} = \frac{1}{-x^2} = -\frac{1}{x^2}$$. 9. **Rewrite the expression:** $$\frac{5}{x} - \frac{1}{x + 1} = \frac{5}{x} + \frac{1}{x^2}$$. 10. **Express in terms of $$x$$:** Recall $$x^2 = -x - 1$$, so: $$\frac{1}{x^2} = \frac{1}{-x - 1} = -\frac{1}{x + 1}$$ (which we already used). But to avoid circular reasoning, use the fact that $$x^3 = 1$$, so: $$\frac{1}{x^2} = x$$ because multiplying both sides by $$x^2$$ gives: $$1 = x^2 \times x = x^3 = 1$$. Therefore: $$\frac{1}{x^2} = x$$. 11. **Final simplification:** $$\frac{5}{x} + \frac{1}{x^2} = 5x^2 + x$$ (since $$\frac{1}{x} = x^2$$ and $$\frac{1}{x^2} = x$$). Using $$x^2 = -x - 1$$: $$5x^2 + x = 5(-x - 1) + x = -5x - 5 + x = -4x - 5$$. **Answer:** $$\boxed{-4x - 5}$$ where $$x$$ is either root $$\frac{-1 + i\sqrt{3}}{2}$$ or $$\frac{-1 - i\sqrt{3}}{2}$$. This completes the solution.