a terrible polynomial problem?

Question

Can anyone of you give me some idea how do I factor out this polynomial, with detail and each stages.
27x^2(3x-y)^2-9x(y-3x)

Answer 1

= 27x²(3x - y)² + 9x(3x - y)
= 9x (3x - y) {(3x(3x - y) + 1}
= 9x (3x - y) {9x² - 3xy + 1}

Answer 2

First factor out 9x:

9x*[3x*(3x-y)^2 -(y-3x)]

Then note that (3x-y)^2 = (y-3x)^2 (reverse sign of the squared term)

this gives 9x*[3x*(y-3x)^2 - (y-3x)] Now factor out the (y-3x):

9x*(y-3x)*[3x*(y-3x) - 1]

Answer 3

Take out the common factor (y-3x)*x

= x(y-3x)*[x(y-3x)-9]