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Let alpha, beta, and gamma be the roots of x³ + a*x² + b*x + c = 0, where a, b, c are real and a != 0. The system of equations alpha*x + beta*y + gamma*z = 0 beta*x + gamma*y + alpha*z = 0 gamma*x + alpha*y + beta*z = 0 has a non-trivial solution. What is the value of a² / b?

1. 1
2. 2
3. 3
4. 4

Correct answer: 3

Solution

Setting the circulant determinant to zero gives alpha³+beta³+gamma³ = 3*alpha*beta*gamma, which by the factorisation identity means alpha+beta+gamma = 0 or alpha = beta = gamma. Since a != 0, we cannot have alpha+beta+gamma = -a = 0; instead alpha = beta = gamma = r, so a = -3r and b = 3r², giving a²/b = 9r²/(3r²) = 3.

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