1. **State the problem:** Given the system of equations: $$x^2 = 2025y + x$$ $$y^2 = 2025x + y$$ Find the value of $$\frac{x}{y^2} + \frac{y}{x^2}$$. 2. **Rewrite the equations:** From the first equation: $$x^2 - x = 2025y$$ From the second equation: $$y^2 - y = 2025x$$ 3. **Express in terms of $$x$$ and $$y$$:** Divide the first equation by $$y^2$$ (assuming $$y \neq 0$$): $$\frac{x^2}{y^2} - \frac{x}{y^2} = 2025 \frac{y}{y^2} = \frac{2025}{y}$$ Divide the second equation by $$x^2$$ (assuming $$x \neq 0$$): $$\frac{y^2}{x^2} - \frac{y}{x^2} = 2025 \frac{x}{x^2} = \frac{2025}{x}$$ 4. **Define variables:** Let $$A = \frac{x}{y^2}$$ and $$B = \frac{y}{x^2}$$. 5. **Rewrite the above in terms of $$A$$ and $$B$$:** From step 3: $$\frac{x^2}{y^2} = x \cdot A$$ because $$\frac{x^2}{y^2} = x \cdot \frac{x}{y^2} = xA$$. Similarly, $$\frac{y^2}{x^2} = y \cdot B$$. So the two equations become: $$xA - A = \frac{2025}{y}$$ $$yB - B = \frac{2025}{x}$$ 6. **Rewrite as:** $$A(x - 1) = \frac{2025}{y}$$ $$B(y - 1) = \frac{2025}{x}$$ 7. **Sum $$A + B$$:** We want to find $$A + B = \frac{x}{y^2} + \frac{y}{x^2}$$. 8. **Try to find a relation between $$x$$ and $$y$$:** Subtract the two original equations: $$x^2 - x - (y^2 - y) = 2025y - 2025x$$ $$x^2 - y^2 - x + y = 2025(y - x)$$ Factor: $$(x - y)(x + y) - (x - y) = 2025(y - x)$$ $$(x - y)(x + y - 1) = 2025(y - x)$$ Note that $$y - x = -(x - y)$$, so: $$(x - y)(x + y - 1) = -2025(x - y)$$ If $$x \neq y$$, divide both sides by $$x - y$$: $$x + y - 1 = -2025$$ $$x + y = -2024$$ 9. **Check if $$x = y$$:** If $$x = y$$, substitute into the first equation: $$x^2 = 2025x + x = 2026x$$ $$x^2 - 2026x = 0$$ $$x(x - 2026) = 0$$ So $$x = 0$$ or $$x = 2026$$. If $$x = y = 0$$, then $$y^2 = 0$$ and denominator zero in the expression, invalid. If $$x = y = 2026$$, check second equation: $$y^2 = 2025x + y$$ $$2026^2 = 2025 \times 2026 + 2026$$ Calculate: $$2026^2 = 2026 \times 2026$$ $$2025 \times 2026 + 2026 = 2026(2025 + 1) = 2026 \times 2026$$ So it holds. 10. **Calculate the expression for $$x = y = 2026$$:** $$\frac{x}{y^2} + \frac{y}{x^2} = \frac{2026}{2026^2} + \frac{2026}{2026^2} = 2 \times \frac{2026}{2026^2} = 2 \times \frac{1}{2026} = \frac{2}{2026} = \frac{1}{1013}$$ 11. **Calculate the expression for $$x + y = -2024$$ and $$x \neq y$$:** From step 8, $$x + y = -2024$$. Rewrite the first original equation: $$x^2 - x = 2025 y$$ Rewrite as: $$x^2 - x - 2025 y = 0$$ Similarly for $$y$$: $$y^2 - y - 2025 x = 0$$ Add these two equations: $$(x^2 + y^2) - (x + y) - 2025(x + y) = 0$$ $$(x^2 + y^2) - (1 + 2025)(x + y) = 0$$ $$(x^2 + y^2) - 2026(x + y) = 0$$ Use $$x + y = -2024$$: $$(x^2 + y^2) - 2026(-2024) = 0$$ $$x^2 + y^2 = 2026 \times 2024$$ Recall: $$(x + y)^2 = x^2 + 2xy + y^2$$ So: $$x^2 + y^2 = (x + y)^2 - 2xy = (-2024)^2 - 2xy$$ Equate: $$(-2024)^2 - 2xy = 2026 \times 2024$$ Calculate: $$2024^2 = 4096576$$ $$2026 \times 2024 = 4101056$$ So: $$4096576 - 2xy = 4101056$$ $$-2xy = 4101056 - 4096576 = 4480$$ $$xy = -2240$$ 12. **Find $$A + B = \frac{x}{y^2} + \frac{y}{x^2}$$:** Rewrite: $$\frac{x}{y^2} + \frac{y}{x^2} = \frac{x^3 + y^3}{x^2 y^2}$$ Use: $$x^3 + y^3 = (x + y)^3 - 3xy(x + y)$$ Calculate numerator: $$(x + y)^3 - 3xy(x + y) = (-2024)^3 - 3(-2240)(-2024)$$ Calculate: $$(-2024)^3 = -2024^3$$ Calculate $$2024^3$$: $$2024^3 = 2024 \times 2024^2 = 2024 \times 4096576 = 8280340224$$ So: $$(-2024)^3 = -8280340224$$ Calculate: $$-3(-2240)(-2024) = -3 \times 2240 \times 2024 = -13595520$$ Sum numerator: $$-8280340224 - 13595520 = -8293935744$$ Calculate denominator: $$(xy)^2 = (-2240)^2 = 5017600$$ So: $$A + B = \frac{-8293935744}{5017600} = -1652.8$$ (approx) 13. **Summary:** - If $$x = y = 2026$$, then $$\frac{x}{y^2} + \frac{y}{x^2} = \frac{1}{1013}$$. - If $$x + y = -2024$$ and $$xy = -2240$$, then $$\frac{x}{y^2} + \frac{y}{x^2} \approx -1652.8$$. Since the problem does not specify which solution, the simplest exact value is for $$x = y = 2026$$: **Final answer:** $$\boxed{\frac{1}{1013}}$$