#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.
