α,β and γ are roots of x3+4x−19=0 Then, α+β+γ=0......(i) αβ+βγ+gα=4.....(ii) αβγ=19.....(iii) From Eq. (ii), α(β+γ)+βγ=4 =α(−α)+βγ=4 [using Eq. (i)] ⇒−α2+
19
α
=4 [using Eq. (iii)] ⇒−α3+19=4α or α3=19−4α This gives
α3
19−4α
=1 ...(iv) Similarly, from Eq. (ii), β(α+γ)+γα=4 ⇒β(−β)+γα=4[using‌Eq.(i)] ⇒−β2+
19
β
=4[using‌Eq.(iii)] ⇒β3=19−4β ⇒
β3
19−4β
=1.....(v) And, from Eq. (ii) γ(α+β)+αβ=4 γ(−γ)+
19
γ
=4 ⇒
γ3
19−4γ
=1......(vi) Adding Eqs.(iv), (v) and (vi), we get