1.

Using the property of determinants and without expanding, prove that:`|1b c a(b+c)1c a b(c+a)1a b x(a+b)|=0`

Answer» `L.H.S.=|{:(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b)):}|=|{:(1,bc,ab+ca+bc),(1,ca,ba+ab+ca),(1,ab,ca+be+ab):}|`
`(C_(3)toC_(3)+C_(2))`
`=(ab+bc+ca)|{:(1,bc,1),(1,ca,1),(1,ab,1):}|`
`=(ab+bc+ca)xx0" "(because C_(1)" and "C_(3)" are same)`
= 0 = R.H.S


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