(d) Let α,β,γ be the roots of the equation x3−5x2−2x+24=0α+β+γ=5⋅⋅⋅⋅⋅⋅⋅(i)αβ+βγ+γα=−2⋅⋅⋅⋅⋅⋅⋅(ii)αβγ=−24⋅⋅⋅⋅⋅⋅⋅(iii)∴αβγ+βγα+γαβ=−24βγ×βγ+−24γα×γα+−24αβ×αβ[using eq.(iii)]=−241[(βγ)2+(γα)2+(αβ)2]=24−1[(αβ+βγ+γα)2−2(αβ2γ+αβγ2+α2βγ)]=24−1[(−2)2−2(−24(α+β+γ))]=24−1[4+2×24×5]=−661