1
Question
If ∣∣
∣
∣∣(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∣∣
∣
∣∣=−k(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−δ), then the value of (k)12
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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∣∣
∣
∣∣=−k(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−δ)letβ+γ−α−δ=a,γ+α−β−δ=b,α+β−γ−δ=c
L.H.S=∣∣
∣
∣∣a4a21b4b21c4c21∣∣
∣
∣∣
=−(a2−b2)(b2−c2)(c2−a2) =−[(β+γ−α−δ)2−(γ+α−β−δ)2][(γ+α−β−δ)2−(α+β−γ−δ)2] [(α+β−γ−δ)2−(β+γ−α−δ)2] =−64(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−δ) ..... [By expanding the terms]
Hence, (k)12=8
L.H.S=∣∣ ∣ ∣∣a4a21b4b21c4c21∣∣ ∣ ∣∣
=−(a2−b2)(b2−c2)(c2−a2)
=−[(β+γ−α−δ)2−(γ+α−β−δ)2][(γ+α−β−δ)2−(α+β−γ−δ)2]
[(α+β−γ−δ)2−(β+γ−α−δ)2]
=−64(α−β)(α−γ)(α−δ)(β−γ)(β−δ)(γ−δ) ..... [By expanding the terms]
Hence, (k)12=8
Hence, (k)12=8
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