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Question

The L.C.M. of v2v, (v1)2, and v31 is .

A
v2(v1)(v2+v+1)
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B
v(v1)3 (v2+v+1)
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C
v(v1)
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D
v (v1)2(v2+v+1)
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Solution

The correct option is D v (v1)2(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 is v(v1)2(v2+v+1).


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