(-i)1/3=∛(-i).
Let z=a+ib=(-i)1/3=∛(-i).
z3=-i=a3+3a2(ib)+3a(ib)2+(ib)3,
z3=-i=a3+3ia2b-3ab2-ib3.
Equate real and imaginary parts:
a3-3ab2=0, a(a2-3b2)=0, a=0 or a=±b√3;
3a2b-b3=-1⇒if a=0, b=1⇒z=(-i)1/3=i;
3a2b-b3=-1⇒if a2=3b2, 9b3-b3=-1, 8b3=-1, b=-½, a2=¾, a=±√3/2⇒z=(-i)1/3=√3/2-i/2 or -√3/2-i/2.
There are 3 cube roots of -i.