LCM of v2−v, (v−1)2, and v3−1
v2(v+1)(v−1) (v2+v+1)
v(v−1)(v−1) (v2+v+1)
v(v+1)(v−1)
v(v−1) (v2+v+1)
The correct option is C v(v−1)(v−1) (v2+v+1)
First expression = v2−v = v(v−1)
Second expression = (v−1)2 = (v−1)(v−1)
Third expression = v3−1 = (v−1) (v2+v+1)
∴ LCM = v(v−1)(v−1) (v2+v+1)
The LCM of v2−v, (v−1)2, and v3−1 is