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Question

LCM of v2v, v212, and v31


A

v2(v+1)(v1) (v2+v+1)

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B

v(v+1)(v1) (v2+v+1)

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C

v(v+1)(v1)

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D

v(v1) (v2+v+1)

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Solution

The correct option is B

v(v+1)(v1) (v2+v+1)


First expression = v2v = v(v1)

Second expression = v212 = (v+1)(v1)

Third expression = v31 = (v1) (v2+v+1)

LCM = v(v+1)(v1) (v2+v+1)


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