Let α, β, γ be the real roots of the equation, x ^ 3 + a x ^ 2 + bx...

Equation ${x}^{3} + a{x}^{2} + bx + c = 0$ has roots $\alpha, \beta, \gamma$. Therefore,

$\alpha + \beta + \gamma = - a$

$\alpha\beta + \beta\gamma + \gamma\alpha = b$

Since the given system of equations has non-trivial solutions,

we have

\left|\begin{cases}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta\end{cases}\right| = 0

or ${\alpha}^{3} + {\beta}^{3} + {\gamma}^{3} - 3\alpha\beta\gamma = 0$

or \left(\alpha + \beta + \gamma\right)\left[{\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} - \alpha\beta - \beta\gamma - \gamma\alpha\right] = 0$</p><p>or \left(\alpha + \beta + \gamma\right)\left[{\left(\alpha + \beta + \gamma\right)}^{2} - 3\left(\alpha\beta + \beta\gamma + \gamma\alpha\right)\right] = 0$

$⇒ - a\left[{a}^{2} - 3b\right] = 0$

or ${a}^{2} / b = 3$