TS EAMCET 04 May 2019 Shift 1 Solved Paper

Consider the equation,
x3−6x2+px+10=0
Now, α,β,γ are in A.P and are roots of the above equation so,
α+β+γ=6‌‌‌‌.......(I)
αβ+βγ+γα=p‌‌‌‌.......(II)
αβγ=−10‌‌‌‌.......(III)
since, α+γ=2β then from equation (I)
3β‌‌=6
β‌‌=2
Substitute 2 for β in equation (III),
αγ=−5
Then from equation (II),
β(α+γ)+γα‌‌=p
2β2+β‌‌=p
p‌‌=8−5
‌‌=3
Now consider
α3+β3+γ3−3αβγ‌‌=(α+β+γ)[

(α+β+γ)2

−3(αβ+βγ+γα)

]
α3+β3+γ3−3(−10)‌‌=6(62−3(3))
α3+β3+γ3‌‌=162−30
‌‌=132