We look at the number of factors of x in each: xy contains a factor of x 1 time. x2 contains a factor of x 2 times. xy3 contains a factor of x 1 time. Therefore the LCM must contain a factor of x the most number of times that any one of them contains x as a factor, which is 2 times. We look at the factors of y (if any) in each: xy contains a factor of y 1 time. x2 contains a factor of y no (0, zero) times. xy3 contains a factor of y 3 times. Therefore the LCD must contain a factor of y the most number of times that any one of them contains y as a factor, which is 3 times. So the LCM contains a factor of x 2 times and a factor of y 3 times: Answer = LCM = Edwin