Question 6201
because everything is being multiplied (and not added or subtracted) you can take every cancellation opportunity:
{{{(3xy^2*2y)/(6Y^4xz^4)}}}
so the (3*2) in the numerator cancels with the 6 in the denominator:
{{{(xy^2*y)/(y^4xz^4)}}}
then the x in the numerator cancels with the x in the denominator:
{{{(y^2*y)/(y^4z^4)}}}
then the (y^2*y) in the numerator will cancel with three of the Y's in the denominator:
{{{1/(yz^4)}}}