First, recall that dividing by a fraction such as #a/b#, is the same as multiplying by #b/a#
#(12w^2+36w+27)/(w-2) div (6w^2+3w-9)/(w-2) = (12w^2+36w+27)/(w-2) times (w-2)/(6w^2+3w-9)#
The w-2 terms cancel out for #w!=2#...
#= (12w^2+36w+27)*1/(6w^2+3w-9) = (12w^2+36w+27)/(6w^2+3w-9)#
To simplify further, we will have to factor the top and bottom. We'll open up with factoring out 3 from the top:
#=(3(4w^2+12w+9))/(6w^2+3w-9)#
Looking at the top, this factors rather quickly into #(2w+3)^2...#
#= (3(2w+3)^2)/(6w^2+3w-9)#
The denominator looks like it will factor into the form of #(ax-3)(bx+3)#. This multiplication would yield #abx^2 -3bx+3ax-9 = abx^2 +3(a-b)-9 = 0#. From this we can determine that #ab=6, a-b = 1# by comparing to our original denominator. Looking at the sets of factors of 6, we'd have the options of #(a=1,b=6), (a=2,b=3), (a=3,b=2),(a=6,b=1) #. Of these, the only one to satisfy #a-b=1# is #a=3, b=2#. Thus we can now factor our denominator...
#(3(2w+3)^2)/(6w^2+3w-9)=(3(2w+3)^2)/((3w-3)(2w+3)#
We can cancel a #2w+3# term on top and bottom, giving...
#= 3(2w+3)/(3w-3)#
Since #3w-3 = 3(w-1)#, we then get..
#=(2w+3)/(w-1)#
