[Klyklops]

You should do your own homework. But I will do it for you this once. The easiest method is to use DeMoivre's Theorem, which presumably you know,

exp(iy) = (cos(y + 2(PI)n) + i sin(y + 2(PI)n)). n is an integer.
put x^3 = a*exp(iy). So y = 0 and a = 192/81.
Then

x = (a^(1/3))*exp(i(y/3))
x = (a^(1/3))*(cos(y/3 + 2(PI)n/3) + i sin(y/3 + 2(PI)n/3))

Here we have
y = 0
so


x = (a^(1/3))*(cos(2(PI)n/3) + i sin(2(PI)n/3))

Then stick in integers for n,

n = 0
x = (a^(1/3))*(cos(0) + i sin(0))
x = (a^(1/3)) = 4/3

n = 1
x = (a^(1/3))*(cos(2(PI)/3) + i sin(2(PI)/3))
x = (2/3)*(1 + i Sqrt(3))

n = 2
x = (a^(1/3))*(cos(4(PI)/3) + i sin(4(PI)/3))
x = (2/3)*(1 - i Sqrt(3))

n >= 3 just cycles through these solutions.

These solutions define an equilateral triangle in the complex plane. PuPu did it a really nice way, and yeah you have to get 3 answers.