I'm stuck with the following:
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12}$
My idea was/is the following:
$\frac{3^3a^6b^9}{5^3}-\frac{c^{12}}{8^2}$
Trouble is that I don't know where to go from here. If I go via LCD for $5^3$ and $8^2$ I'll get 8000 in the denominator and high numbers in the numerator, which I think isn't the point here with factorization. Ideally I'd get something like $(3a^3b^3)^2$ in the first numerator but for the whole expression, of course. Maybe I'm looking at it for too long, so I've lost perspective, but usual helping tools like wolfram and symbolab aren't of much help.
Use $x^3-y^3=(x-y)(x^2+xy+y^2)$
$\frac{27}{125}a^6b^9-\frac{1}{64}c^{12} =\left(\frac{3a^2b^3}{5}\right)^3-\left(\frac{c^4}{4}\right)^3$.
Now, let $x=\frac{3a^2b^3}{5}$ and $y=\frac{c^4}{4}$ and use that factorization to get $[\frac{3a^2b^3}{5}-\frac{c^4}{4}][\frac{9a^4b^6}{25}+\frac{3a^2b^3c^4}{20}+\frac{c^8}{16}]$