Let α ,β ,γ be the real roots of the equation, x^3+ax^2+bx+c=0, (a,b,c∈ R and a,b≠ 0). If the system of equations (in, u,v,w) given by α u+β v+γ…

Equation ${x}^{3}+a{x}^{2}+bx+c=0$x3+ax2+bx+c=0 has roots $\alpha ,\beta ,\gamma$α,β,γ. Therefore,

$$\alpha +\beta +\gamma =-a$$α+β+γ=−a

$$\alpha \beta +\beta \gamma +\gamma \alpha =b$$αβ+βγ+γα=b

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

we have

$$|\begin{matrix}\alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{matrix}|=0$$∣αβγ​βγα​γαβ​∣=0

or ${\alpha }^{3}+{\beta }^{3}+{\gamma }^{3}-3\alpha \beta \gamma =0$α3+β3+γ3−3αβγ=0

or $(\alpha +\beta +\gamma )[{\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}-\alpha \beta -\beta \gamma -\gamma \alpha ]=0$(α+β+γ)[α2+β2+γ2−αβ−βγ−γα]=0

or $(\alpha +\beta +\gamma )[{(\alpha +\beta +\gamma )}^{2}-3(\alpha \beta +\beta \gamma +\gamma \alpha )]=0$(α+β+γ)[(α+β+γ)2−3(αβ+βγ+γα)]=0

$$\Rightarrow -a[{a}^{2}-3b]=0$$⇒−a[a2−3b]=0

or ${a}^{2}/b=3$a2/b=3