Once you have used your difference of cubes to factor

8x^{3}-27=(2x-3)(4x^{2}+6x+9)

you know that x=3/2 is a solution

Then you need to find the other solutions. If we only cared about real roots we would use the discriminant to see if it was worth solving the quadratic, but since we want complex roots as well we can just go ahead and commit to using the quadratic formula for the remaining 4x^{2}+6x+9

a=4 b=6 c=9

(-6±√(36-4(4)(9)))/8

(-6±√(36-144))/8

(-6±√(-108))/8

√(-108)=*i*√(108)=*i*√(3*36)=6*i*√(3)

(-6±6*i*√(3))/8

(-3±3*i*√(3))/4

So our roots are one real root x=3/2 and two complex roots x=-3/4+3*i*√(3)/4 and x=-3/4-3*i*√(3)/4