The difference of cubes formula states that
#a^3 - b^3 = (a-b)(a^2+ab+b^2)#
(try expanding the right hand side to verify this)
Applying the above formula, we have
#27-8t^3 = 3^3-(2t)^3#
#=(3-2t)(3^2+3(2t)+(2t)^2)#
#=(-2t+3)(4t^2+6t+9)#
If we are limiting ourselves to the reals, then we are done, as the discriminant #6^2-4(4)(9)# of #4t^2+6t+9# is less than zero. If we are willing to use complex numbers, then by the quadratic formula, #4t^2+6t+9# has the roots
#t = (-6+-sqrt(-108))/8#
#=(-6+-6sqrt(3)i)/8#
#=-3/4+-(3sqrt(3))/4i#
and so, multiplying by #4# to obtain the correct coefficient for #t#, we have the complete factorization as
#27-8t^3 = (-2t+3)(2t + 3/2 -(3sqrt(3))/2i)(2t+3/2+(3sqrt(3))/2i)#
