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" मान Fिकों "A=|[1,bc,a(b+c)],[1,ac,b(c+...

" मान Fिकों "A=|[1,bc,a(b+c)],[1,ac,b(c+a)],[1,ab,c(a+b)]|

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|[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))|=

|(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

Prove the following : [[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, prove that : |{:(1, bc, a(b+ c) ),(1, ca, b ( c+ a) ),(1, ab , c( a+ b)):}|=0 .