1. **State the problem:** Given the equation $$a \cos^{12} A + b \cos^{8} A + c \cos^{6} A - 1 = 0$$ and the condition $$\sin A + \sin^2 A = 1,$$ find the value of $$\frac{b+c}{a+b}.$$ 2. **Analyze the given condition:** From $$\sin A + \sin^2 A = 1,$$ let $$x = \sin A.$$ Then we have: $$x + x^2 = 1.$$ Rearranging: $$x^2 + x - 1 = 0.$$ 3. **Solve for $$\sin A$$:** Using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}.$$ Since $$\sin A$$ must be between -1 and 1, the valid root is: $$\sin A = \frac{-1 + \sqrt{5}}{2}.$$ 4. **Find $$\cos^2 A$$:** Using the Pythagorean identity: $$\cos^2 A = 1 - \sin^2 A.$$ Calculate $$\sin^2 A$$: $$\sin^2 A = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}.$$ Then: $$\cos^2 A = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2}{2} - \frac{3 - \sqrt{5}}{2} = \frac{2 - 3 + \sqrt{5}}{2} = \frac{-1 + \sqrt{5}}{2}.$$ 5. **Express powers of $$\cos A$$:** Note that: $$\cos^2 A = \frac{-1 + \sqrt{5}}{2}.$$ Then: $$\cos^6 A = (\cos^2 A)^3 = \left(\frac{-1 + \sqrt{5}}{2}\right)^3,$$ $$\cos^8 A = (\cos^2 A)^4 = \left(\frac{-1 + \sqrt{5}}{2}\right)^4,$$ $$\cos^{12} A = (\cos^2 A)^6 = \left(\frac{-1 + \sqrt{5}}{2}\right)^6.$$ 6. **Substitute into the original equation:** Let $$y = \cos^2 A = \frac{-1 + \sqrt{5}}{2}.$$ Then: $$a y^6 + b y^4 + c y^3 - 1 = 0.$$ 7. **Rewrite the equation:** $$a y^6 + b y^4 + c y^3 = 1.$$ 8. **Goal:** Find $$\frac{b+c}{a+b}.$$ 9. **Note:** Without additional information or relations between $$a, b, c,$$ this expression cannot be simplified further from the given data alone. The problem likely expects recognizing a pattern or using the equation to express $$a, b, c$$ in terms of each other. 10. **Assuming the equation holds for $$y$$, and considering the polynomial form, one approach is to consider the polynomial as a function of $$y^3$$:** Let $$z = y^3,$$ then: $$a z^2 + b y^{4} + c z = 1.$$ But since $$y^4 = y \cdot y^3 = y z,$$ the equation becomes: $$a z^2 + b y z + c z = 1.$$ 11. **Factor out $$z$$:** $$a z^2 + z (b y + c) = 1.$$ 12. **Rewrite as:** $$a z^2 + (b y + c) z - 1 = 0.$$ 13. **Since $$z = y^3$$ is a root, this quadratic in $$z$$ must be satisfied.** 14. **Without more constraints, the problem is underdetermined.** **Final answer:** The value of $$\frac{b+c}{a+b}$$ cannot be determined uniquely from the given information.