first factor #(x^2-9)# since it's a difference of two squares. then factor out #6x+18# so at this point, you have: #(5x)/((x-3)(x+3))*(6(x+3))/(15x^3)# now you see that you can cancel #(x+3)# since there's two of them. now you can divide #5# into #15# but now you can divide #3# into the #6# on top of it. at this point, you have #x/(x-3)*2/(x^3)# as you can see, you can take out one #x# from #x^3# so that will only leave you with your answer