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([23,1,1,bc,bc(b+c)],[1,ca,ca(c+a)],[1,a...

([23,1,1,bc,bc(b+c)],[1,ca,ca(c+a)],[1,ab,ab(a+b)]|=0

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|[1,bc,a(b+c)],[1,ca,b(c+a)],[1,ab,c(a+b)]|=0

|(1,bc,bc(b+c)),(1,ca,ca(c+a)),(1,ab,ab(a+b))|=

det[[1,bc,bc(b+c)1,ca,ca(c+a)1,ab,ab(a+b)]]=0

|{:(1,bc,bc(b+c)),(1,ca,ca(c+a)),(1,ab,ab(a+b)):}|=0

Without expanding show that : |(1,bc,bc(b+c)),(1,ca,ca(c+a)),(1,ab,ab(a+b))|=0

Using the properties of determinants,prove det[[1,bc,bc(b+c)1,ca,ca(c+a)1,ab,ab(a+b)]]

Prove the following : [[1,bc,a(b+c)],[1,ca,b(c+a)],[1,ab,c(a+b)]] =0

|(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b))|= 0

|(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b))|=