Q. 2 E4.4( 7 Votes )

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Answer :

In the given term


Dividend = 15 (a3b2c2– a2b3c2 + a2b2c3)


Take out the common part in binomial term


= 3 × 5(a × a × a × b × b × c × c– a × a × b × b × b × c × c + a × a × b × b × c × c × c )


= 15 a2b2c2(a – b + c)


Divisor = 3abc


=


= 5abc[a-b + c]


= [5a2bc– 5ab2c + 5abc2]


Hence dividing 15 (a3b2c2– a2b3c2 + a2b2c3) by 3abc gives out [5a2bc– 5ab2c + 5abc2]


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