First, divide the numbers, the #a#'s and the #b#'s:
#(10a^3b^3)/(15ab^8) = (10 * a^3 * b^3)/(15 * a * b^8) = (10/15) * (a^3/a) * (b^3/b^8)#
Now, #10/15# you can cancel since #10 = 5 * 2# and #15 = 5 * 3#, so
#10/15 = (cancel(5)*2)/(cancel(5)*3) = 2/3#.
With #a# and #b#, you can also cancel:
#a^3/a = (a * a * cancel(a)) / cancel(a) = (a * a) / 1 = a * a = a^2#
#b^3/b^8 = (b * b * b) / (b * b * b * b * b * b * b * b) = (cancel(b * b * b)) / (b * b * b * b * b * cancel(b * b * b)) = 1 / (b * b * b * b * b) = 1/ b^5#
Alternatively, remember the power rules:
#1/x = x^(-1)#
#x^a * x^b = x^(a+b)#
#x^a / x^b = x^(a-b)#
In this case, they can be used as follows:
# a^3 / a = a^(3-1) = a^2#
#b^3 / b^8 = b^(3-8) = b^(-5) = 1/(b^5)#
So, in total you have:
#(10a^3b^3)/(15ab^8) = (10/15) * (a^3/a) * (b^3/b^8) = 2/3 * a^2 * 1/(b^5) = (2a^2)/(3b^5)#
