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Question

The LCM of v2v, (v1)2, and v31 is


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' can be written as 'v(v1)'.
Second expression, '(v1)2' can be written as '(v1)(v1)'.
Third expression = v31 = (v1) (v2+v+1)

So, their LCM = v(v1)2(v2+v+1)


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