1.

The value of [(a-b)(b-c)(c-a)] is equal to

Answer»

0
1
2[ a b c ]
2

Solution :Now, `[(a-b)(b-c)(c-a)]`
`= (a-b).[(b-c) XX (c-a)]`
`= (a-b).[bxxc-bxxa-cxxc+cxxa]`
`= (a-b).[bxxc-bxxa+cxxa] ""[ :' c xx c = 0]`
`= a.[bxxc]- a.[bxxa] + a.[cxxa] - b.[bxxc]+b.[bxxa] - b.[cxxa]`
`= [abc] - 0 + 0 + 0 - [BCA]`
`= [abc] - [abc] = 0`


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