So we have
#((-15(ab)^2)/(5a^2b))^5#
First let's take that first square in the first term, since it's a multiplication the exponent goes to both numbers.
#((-15a^2b^2)/(5a^2b))^5#
Now we can get rid of any letters that show up equally in top and bottom, we have 2 #a#s on top and 2 in the bottom so they both go, but we only 1 #b# in the bottom and two on the top, so one of the #b#s on the top will remain.
#((-15b)/5)^5#
Now, we simplify the actual numbers, we know #15 = 3*5# so it's just a matter of factorizing and discarding equal factors on top and bottom.
#(-3b)^5#
Which is easy to compute if you so wish, the letter just gets the exponent, the minus sign continues to be a minus because #(-1)\*(-1)^2*(-1)^2 = (-1)*1*1=-1#, which only leaves us to deal with the #3#, but since #3^5 = 3*(3)^2*(3)^2=3*9*9=3*81#, we only need to do one multiplication that isn't tabled.
#3*81 = 3*80 + 3*1 = 243#, so the final expression is
#(-3b)^5# or #-243b^5#
