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B
2a2b2c+6a2b2c2
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C
2a2bc+6ab2c
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D
2ab2c+6b2c2
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Solution
The correct option is B2a2b2c+6a2b2c2 We use distributive law to multiply a monomial and a binomial, as shown below: (2abc)(ab+3cab) =(2abc×ab) + (2abc×3cab) =2a2b2c+6a2b2c2