Ok here it goes!
$$\frac{12a^5b^3(3-b)}{4a^4b(b^2+b-12)}\\\\=\frac{3a^5b^3(3-b)}{1a^4b(b^2b-12)}\mbox{I\;have\;canceled\;top\;and\;bottom\;by\;4.}\\\\=\frac{3aaaaabbb(3-b)}{4aaaab(bb+b-12)}\mbox{Now\;that\;I\;have\;expanded,\;there\;are\;no\;invisible\;signs\;between\;the\;letters.}\\\\=\frac{3abbb(3-b)}{4b(bb+b-12)}\mbox{I\;have\;cancelled\;out\;}aaaa\;\mbox{top\;and\;bottom.}\\\\=\frac{3abb(3-b)}{4(bb+b-12)}\mbox{I\;have\;cancelled\;}b\;\mbox{top\;and\;bottom.}\\\\=\frac{3abb(3-b)}{4(bb+b-12)}$$
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