If α,β are the roots of ax2+bx+c=0(a≠0) and α+β,α2+β2,α3+β3 are in G.P., then the value of c⋅Δ is (where Δ=b2−4ac)
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Solution
ax2+bx+c=0 ⇒αβ=ca
α+β,α2+β2,α3+β3 are in G.P. ⇒(α+β)×(α3+β3)=(α2+β2)2⇒β3α+βα3=2α2β2⇒β3α−α2β2−α2β2+βα3=0⇒β2α(β−α)−βα2(β−α)=0⇒(β−α)(βα2−αβ2)=0⇒αβ(α−β)2=0⇒αβ=0 or α=β⇒c=0 or Δ=0∴c⋅Δ=0