Question

# The H.C.F. of 3xy2z3, 3yz2x2 and x2y2z is .x2yz3xy2z3xyzx2y2z

Solution

## The correct option is C xyzIn the three given quantities, the common factors are x, y and z. The highest powers of x, y and z that exactly divide the given quantities are x1,y1 and z1 respectively.

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