If $z^2 + z + 1 = 0$, where $z$ is complex number, then the value of $(z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + ...+ (z^6 + \frac{1}{z^6})^2$ is :

Question

( A ) 18
( B ) 54
( C ) 12
( D ) 6