1. We are given a cubic equation $$x^3 + ax^2 + bx + c = 0$$ with roots $$\alpha, \beta, \gamma$$ and the system: $$\alpha u + \beta v + \gamma w = 0,$$ $$\beta u + \gamma v + \alpha w = 0,$$ $$\gamma u + \alpha v + \beta w = 0$$ which has a non-trivial solution $$(u, v, w) \neq (0,0,0)$$. We need to find the value of $$\frac{a^2}{b}$$. 2. The system matrix is: $$\begin{pmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{pmatrix}$$ Non-trivial solutions exist if and only if the determinant of this matrix is zero. 3. Denote the matrix as $$M$$, then $$\det(M) = 0$$. 4. By calculating or using symmetry properties, we find: $$\det(M) = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$$ 5. Since $$\alpha, \beta, \gamma$$ are roots of $$x^3 + ax^2 + bx + c = 0$$: - $$\alpha + \beta + \gamma = -a$$ - $$\alpha\beta + \beta\gamma + \gamma\alpha = b$$ - $$\alpha \beta \gamma = -c$$ 6. Substitute: $$\det(M) = (-a)(\alpha^2 + \beta^2 + \gamma^2 - b) = 0$$ 7. Use the identity: $$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (-a)^2 - 2b = a^2 - 2b$$ 8. Substitute back: $$-a (a^2 - 2b - b) = -a(a^2 - 3b)=0$$ 9. For non-trivial solution, either: - $$a=0$$ (contradicts condition $$a \neq 0$$) - OR $$a^2 - 3b=0 \Rightarrow a^2 = 3b$$ 10. Therefore: $$\frac{a^2}{b} = 3$$ Final answer: 3