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)
