Question 161452
When you see:
x^3/x^2
You could rewrite it as:
(x*x*x)/(x*x)
Now, for each matching pair of x (one in the numerator and one in the denominator) you can "cancel" it out.  Since, there are two pairs, you can cancel out two x's in the numerator and two x's in the denominator leaving:
x/1
or 
x
.
.
Similarly, for your:
x^3y^2/5xy^6 
Looking at just the x's:
(x*x*x)y^2/5xy^6
Notice, there is only ONE pair of x's so canceling them out gives:
(x*x)y^2/5y^6
or
x^2y^2/5y^6
.
Looking at just the y's:
x^2(y*y)/5(y*y*y*y*y*y)
Since there are two pairs of y's, we can cancel two from the numerator and two from the denominator:
x^2/5(y*y*y*y)
or
x^2/(5y^4)