The correct option is B 18(2x−1)2(x−3)
Using the identity
(a−b)2=a2−2ab+b2,
We can write
(4x2−4x+1)=(2x−1)2
Hence, 6(4x2−4x+1)=2× 3× (2x−1)2.
Also, (2x2−7x+3)=(2x2−6x−x+3)=2x(x−3)−1(x−3)=(2x−1)(x−3).
Hence, 9(2x2−7x+3)=32(2x−1)(x−3).
So, the irreducible factors occuring in 6(4x2−4x+1) and 9(2x2−7x+3) are 2,3,(2x−1)and(x−3)
The greatest exponents of these factors respectively are 1,2,2 and 1.
Hence, the LCM = 21× 32× (2x−1)2× (x−3)1=18(2x−1)2(x−3).