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Question

Using the property of determinants and without expanding, prove that

∣ ∣ ∣1bca(b+c)1cab(c+a)1abc(a+b)∣ ∣ ∣=0

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Solution

Let Δ=∣ ∣ ∣1bca(b+c)1cab(c+a)1abc(a+b)∣ ∣ ∣
=∣ ∣1bcab+ac1cabc+ab1abca+bc∣ ∣
Aplying R1=R1R2&R2=R2R3
Δ=∣ ∣0bccaacbc0caababca1abc(a+b)∣ ∣
Taking (1 common from C2 we get,
Δ=∣ ∣0acbcacbc0abcaabca1abc(a+b)∣ ∣=0
Since R1 and R2 are same.

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