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B
12a+6a2+3a3−412
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C
12a+6a2+3a3−2424
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D
12a+6a2+3a3−424
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Solution
The correct option is D12a+6a2+3a3−424 a2+a24+a38−16 First find the LCM of denominator, 2,4,8 and 6 So, the LCM of 2,4,6 and 8 is 24 Therefore, 12a+6a2+3a3−424