1
Question
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) −γ3(γ−α)(γ−β)(γ−δ) =1−∑α3(α−β)(α−γ)(α−δ)
Hence ∑α3(α−β)(α−γ)(α−δ)=1−δ3(δ−α)(δ−β)(δ−γ)
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)
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)
−γ3(γ−α)(γ−β)(γ−δ) =1−∑α3(α−β)(α−γ)(α−δ)
Hence ∑α3(α−β)(α−γ)(α−δ)=1−δ3(δ−α)(δ−β)(δ−γ)
Hence ∑α3(α−β)(α−γ)(α−δ)=1−δ3(δ−α)(δ−β)(δ−γ)
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