TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the equation.
x3−9x2+26x−24=0
The roots are such that α=2β
So,
α+β+γ=9
αβ+βγ+γα=−26
And,
αβγ=24
Therefore,
3β+γ=9
And,
β2γ=12
Now,
β2(9−3β)=12
β3−3β2+4=0
(β+1)(β2−4β+4)=0
(β+1)(β−2)2=0
Therefore,
β=−1,2
When β=−1 then α=−2
Then,
α3+β3=(−2)3+(−1)3
=−8−1
=−9
When β=2 then α=4.
Then,
α3+β3=(4)3+(2)3
=64+8
=72