2213
Question
Prove that ∣∣
∣
∣∣(β+γ−α−δ)4(β+γ−α−δ)21(γ+α−β−δ)4(γ+α−β−δ)21(α+β−γ−δ)4(α+β−γ−δ)21∣∣
∣
∣∣=−64(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−δ)
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Solution
∣∣
∣
∣∣(β+γ−α−δ)4(β+γ−α−δ)21(γ+α−β−δ)4(γ+α−β−δ)21(α+β−γ−δ)4(α+β−γ−δ)21∣∣
∣
∣∣
Let β+γ−α−δ=X,γ+α−β−δ=Y,α+β−γ−δ=Z
∴
∣∣
∣
∣∣X4X21Y4Y21Z4Z21∣∣
∣
∣∣
=R1→R1−R2,R2→R2−R3
=∣∣
∣
∣∣X4−Y4X2−Y20Y4−Z4Y2−Z20Z4Z21∣∣
∣
∣∣
=(X4−Y4)(Y2−Z2)−(X2−Y2)(Y4−Z4)+0
=X4Y2−X4Z2−Y6+Y4Z2−[X2Y4−X2Z4−Y6+Y2Z4]
=X4Y2−X4Z2−Y6+Y4Z2−X2Y4+X2Z4+Y6−Y2Z4
=X2Y2(X2−Y2)+X2Z2(Z2−X2)+Y2Z2(Y2−Z2)
=(β+γ−α−δ)2(γ+α−β−δ)2[(β+γ−α−δ)2−(γ+α−β−δ)2]+(β+γ−α−δ)2(α+β−γ−δ)2[(α+β−γ−δ)2−(β+γ−α−δ)2]+(γ+α−β−δ)2(α+β−γ−δ)2[(γ+α−β−δ)2−(α+β−γ−δ)2]
Solving the above equation, we get,
=−64(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−α)
∣∣
∣
∣∣(β+γ−α−δ)4(β+γ−α−δ)21(γ+α−β−δ)4(γ+α−β−δ)21(α+β−γ−δ)4(α+β−γ−δ)21∣∣
∣
∣∣
Let β+γ−α−δ=X,γ+α−β−δ=Y,α+β−γ−δ=Z
∴
∣∣
∣
∣∣X4X21Y4Y21Z4Z21∣∣
∣
∣∣
=R1→R1−R2,R2→R2−R3
=∣∣
∣
∣∣X4−Y4X2−Y20Y4−Z4Y2−Z20Z4Z21∣∣
∣
∣∣
=(X4−Y4)(Y2−Z2)−(X2−Y2)(Y4−Z4)+0
=X4Y2−X4Z2−Y6+Y4Z2−[X2Y4−X2Z4−Y6+Y2Z4]
=X4Y2−X4Z2−Y6+Y4Z2−X2Y4+X2Z4+Y6−Y2Z4
=X2Y2(X2−Y2)+X2Z2(Z2−X2)+Y2Z2(Y2−Z2)
=(β+γ−α−δ)2(γ+α−β−δ)2[(β+γ−α−δ)2−(γ+α−β−δ)2]+(β+γ−α−δ)2(α+β−γ−δ)2[(α+β−γ−δ)2−(β+γ−α−δ)2]+(γ+α−β−δ)2(α+β−γ−δ)2[(γ+α−β−δ)2−(α+β−γ−δ)2]
Solving the above equation, we get,
=−64(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−α)
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