The sum of non-real roots of the polynomial equation x3 + 3x2 + 3x + 3 = 0.\t1.lies between 0 and 12.equals 03.lies between -1 and 04.has absolute value bigger than 1

Let the the roots be a,b,c....in which b and c be the 2 non real roots...we need to find b+c...On the other hand we have:::a+b+c = -3....ab+bc+ca = 3...abc = -3...Now we use these 3 equations...bc = -3/a...putting this in 2nd eq...we get....-> a(b+c) - 3/a = 3-> a² (b+c) - 3a - 3 = 0...which is a quadratic in `a`..Discriminant of this should be 0 because only one possible value of a can be root of the given cubic equation...hence on doing so...-> 9 + 12(b+c) = 0-> (b+c) = -3/4....ans....option 3...