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]
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Questions on Factorization41 mins
NCERT | Factorisation Problems38 mins
Quiz | Important Questions on Factorisation35 mins
Algebraic Identities - Mark your Weak Areas41 mins
Champ Quiz | Division of Polynomials33 mins
Quiz | Questions on Factorization41 minsNCERT | Factorisation Problems38 minsQuiz | Important Questions on Factorisation35 minsAlgebraic Identities - Mark your Weak Areas41 minsChamp Quiz | Division of Polynomials33 minsTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
RELATED QUESTIONS :
<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics<span lang="EN-US
AP - Mathematics