Prove that ∑α3(α−β)(α−γ)(α−δ)=1−δ3(δ−α)(δ−β)(δ−γ).
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Solution
We have, by actual division x3(x−α)(x−β)(x−γ)=1+f(x)(x−α)(x−β)(x−γ) Since 11.1.1=1 where f(x) is polynomial of second degree. Let x3(x−α)(x−β)(x−γ)=1+Ax−α+Bx−β+Cx−γ ...(i) ∴A=limx→α(x−α)[x3(x−α)(x−β)(x−γ)]=limx→α[x3(x−β)(x−γ)]=α3(α−β)(α−γ) B=limx→β(x−β)[x3(x−α)(x−β)(x−γ)]=limx→β[x3(x−α)(x−γ)]=β3(β−α)(β−γ) and C=limx→γ(x−γ)[x3(x−α)(x−β)(x−γ)]=limx→γ[x3(x−α)(x−β)]=γ3(γ−α)(γ−β) Substituting the values of A, B, C in (i), we have x3(x−α)(x−β)(x−γ)=1+α3(α−β)(α−γ)(x−α)+β3(β−α)(β−γ)(x−β)
+γ3(γ−α)(γ−β)(x−γ) which are required partial fractions. ∴x3(x−α)(x−β)(x−γ)=1−α3(α−β)(α−γ)(α−x)−β(β−α)(β−γ)(β−x)