Consider the cubic equations. x3−ax2+bx−c=0 The sum of the roots of the equation is α+β+γ=a And αβ+βγ+γα=b The product of the roots is αβγ=c Therefore, ∑α2(β+γ)=∑α2(a−α) =a∑α2−∑α3 =[
a[(α+β+γ)2−2(αβ+βγ+γα)]−
[(α+β+γ)(α2+β2+γ2−αβ−βγ−γα)]+
3αβγ
] =a[a2−2b]−[a(a2−3b)+3c] Solve further ∑α2(β+γ)=ab−3c