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Question

Prove that

∣ ∣ ∣ ∣1ab1a+1b1bc1b+1c1ca1c+1a∣ ∣ ∣ ∣=0

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Solution

1a+1b=a+bab=c(a+b)abc

=1abc

∣ ∣1abbc+ca1bcca+ab1caab+bc∣ ∣ =0

by C3+C2 and taking (ab) out thus making C1 and C3 identical.

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