If α,β,γ are roots of the cubic x3−2x+3=0, then the value of 1α3+β3+6+1β3+γ3+6+1γ3+α3+6 equals to

Since α,β,γ are the roots of x3−2x+3=0……(1) Then S1=α+β+γ=0,S2=αβ+βγ+γα=−2,S3=αβγ=−3 Substituting α,β in equation (1), we get α3=2α−3,β3=2β−3∴α3+β3=2(α+β)−6⇒α3+β3+6=2(α+β)=2-γ∴∑1α3+β3+6=∑−12γ=−121α+1β+1γ=−12αβ+βγ+γααβγ=−12−2−3=−13