Question:

The system of equations is: \alpha u + \beta v + \gamma w = 0, \beta u + \gamma v + \alpha w = 0, \gamma u + \alpha v + \beta w = 0. For a non-trivial solution, the determinant of the coefficient matrix must be zero: |\alpha \beta \gamma; \beta \gamma \alpha; \gamma \alpha \beta| = 0. This simplifies to -(\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma) = 0, which means \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma. Since \alpha, \beta, \gamma are roots of x^3 + ax^2 + bx + c = 0, we have \alpha+\beta+\gamma = -a, \alpha\beta+\beta\gamma+\gamma\alpha = b, \alpha\beta\gamma = -c. Using the identity \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)((\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)), we get -a(a^2 - 3b) = 0. Since a \neq 0, we have a^2 - 3b = 0, so a^2/b = 3.