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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