If α,β are the roots of ax2+bx+c=0 then (aα+b)−3 +(aβ+b)−3
(b3 – 3abc)/ a3c3
α+β= −ba , αβ=ca
α ≠ 0 is root of the equation ax2+bx+c =0 then aα2+bα+c=0 => aα+b=ca
similarly αβ+b = −cβ
so (aα+b)−3 +(aβ+b)−3 = −(α3+β3)c3 = −[(α+β)3−3αβ(α+β)]c3 = b3−3abca3c3