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Question

From the sum of 3a2b+4c and 3b2c subtract abc.

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Solution

The value of terms as per the question is calculated as shown below
(3a2b+4c)+(3b2c)(abc)
=3a2b+4c+3b2ca+b+c
On further calculation, we get
=3aa+3b+b2b+4c+c2c
=2a+2b+3c
Hence, (3a2b+4c)+(3b2c)(abc)=2a+2b+3c

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