Math, asked by Dhruvirohit, 2 months ago

On factorising (3a^2bc+9ab^2c+21abc^2)​

Answers

Answered by sejwalhimanshu1
2

Step-by-step explanation:

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Answered by alekhyaapati
1

Answer:

Answer

30(a

2

bc+ab

2

c+abc

2

)

=2×3×5[(a×a×b×c)+(a×b×b×c)+(a×b×c×c)]

=2×3×5×a×b×c(a+b+c)

Thus 30(a

2

bc+ab

2

c+abc

2

)÷6abc

=

2×3×abc

2×3×5×abc(a+b+c)

=5(a+b+c)

Alternatively 30(a

2

bc+ab

2

c+abc

2

)÷6abc

=

6abc

30a

2

bc

+

6abc

30ab

2

c

+

6abc

30abc

2

=5a+5b+5c

=5(a+b+c)

Step-by-step explanation:

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