Dont bring in powers of TWO because since u already have a pwoer of THREE, and a 1, u can use cubic factorisation
If u did it ur way:
z^3 = -i^2
z^3 + i^2 = 0
How do u solve this?
Go back,
z^3 = -i^2
If u REALLY wanna keep the i2, then let z = rcis(theta) to solve for 'z'