For sum of coefficients in the expansion of (αx2−2x+1)35 we substitute x=1
We get sum of coefficients of (αx2−2x+1)35
as (α−2+1)35
=(α−1)35 ...(i)
Similarly for sum of coefficients of (x−αy)35 we substitute x=1 and y=1
We get sum of coefficients (1−α)35 ...(ii)
It is given that the sum of coefficients are equal.
Hence (1−α)35=(α−1)35
α=1