Notice that #27#, #x^3# and #1/8# are all cubes. If you were allowed complex numbers as coefficients, then this could be completely factored as #(3x - 1/2)(3 omega x - 1/2)(3 omega ^2x - 1/2)#, where #omega# is the complex cube root of 1.

As it is, you are only interested in real numbers, not complex ones. So the only linear factor is #(3x - 1/2)#. The other factor is #(9x^2+(3/2)x+1/4)#.

Here are a couple of ways of finding that quadratic factor:

(1) Using complex arithmetic:

#(3 omega x - 1/2)(3 omega ^2x - 1/2) = 9 omega^3 x^2 - (3/2)(omega + omega^2)x + 1/4 = 9x^2 + (3/2)x + 1/4#

(since #omega^3 = 1# and #omega + omega^2 = -1#).

(2) Using real arithmetic, solve:

#27x^3 - 1/8 = (3x - 1/2)(ax^2 + bx + c)#

#= 3ax^3 + (3b - (1/2)a)x^2 + (3c - (1/2)b)x - (1/2)c#

Comparing coefficients, we get:

#3a = 27#

#(3b - (1/2)a) = 0#

#(3c - (1/2)b) = 0#

#(1/2)c = 1/8#

yielding #a = 9#, #b = 3/2# and #c = 1/4#.