Given that α, β, a, b are in A.P.
⇒b−α=3(β−α)⇒b=3β−2α....(1)
Also, α, β, c, d are in G.P.
⇒dα=(βα)3⇒d=β3α2.....(2)
Similarly, using the last condition, we get
1f=3β−2α......(3)
Given that b,d,f are G.P.,
d2=bf⇒β6α4=αβ(3β−2α3α−2β)⇒β5(3α−2β)=α5(3β−2α)⇒2(α6−β6)=3αβ(α4−β4)
So, 2(α6−β6)αβ(α4−β4)=3