8y^3-27 = (2y)^3 - 3^3
=(2y-3)((2y)^2+3(2y)+3^2)
=(2y-3)(4y^2+6y+9)
based on the difference of cubes identity:
m^3-n^3 = (m-n)(m^2+mn+n^2),
with m=2y and n=3.
Note that (4y^2+6y+9) has no simpler factors with real coefficients. To check this, note that it is of the form ay^2+by+c with a=4, b=6 and c=9, which has discriminant given by the formula:
Delta = b^2-4ac
= 6^2-(4xx4xx9) = 36 - 144 = -108 < 0
Just as a potential point of interest, if you were allowed complex coefficients, this would factor as:
(4y^2+6y+9) = (2y-3omega)(2y-3omega^2)
where omega = -1/2 + sqrt(3)/2i is known as the primitive cube root of unity.
