Let α , β , γ be the real roots of the equation, x 3 + a x 2 + b x + c = 0 , ( a , b , c…

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 

$\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{array}\right|=0$

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

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

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

$\Rightarrow -a\left[a^2-3b\right]=0$

or $a^2/b=3$