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Question

# Subtract: 3a(a+b+c)−2(ab+c) from 4c(a+b+c)

A
3a22b2+4c2ab+6bc+7ac
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B
3a2+2b2+4c2ab+6bc7ac
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C
ac+4bc+4c23a2ab+2c
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D
ac+4bc+4c2+3a2ab+2c
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Solution

## The correct option is D −3a2+2b2+4c2−ab+6bc−7ac3a(a+b+c)−2b(a+b+c)⇒3a(a+b+c)−2b(a+b+c)=3a2+3ab+3ac−(2ab+2b2+2bc)⇒3a(a+b+c)−2b(a+b+c)=3a2+3ab+3ac−2ab−2b2−2bc⇒3a(a+b+c)−2b(a+b+c)=3a2−2b2+ab+3ac−2bcRequired difference ⇒4c(−a+b+c)−(3a2−2b2+ab+3ac−2bc)⇒4c(−a+b+c)−(3a2−2b2+ab+3ac−2bc)=−4ac+4bc+4c2−3a2+2b2−ab−3ac+2bc⇒4c(−a+b+c)−(3a2−2b2+ab+3ac−2bc)=−3a2+2b2+4c2−7ac+6bc−ab

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