The simplest way to explain/to illustrate it is to use square roots of complex number.

    For any complex number z, if  =  is a square root of z, then  is also square root of z  (the second value),   

    and  = 0.

    -i =  = .  

    Therefore, the three values of the cube root of -i are

        a)   =  = i;

        b)   =  = ;

        c)   =  = .

        Therefore, the sum of the three cube roots is

         =  = 0,

        since  = 0,  as you can easily check.

    You can check it on your own as an exercise for you.

She used trigonometry, but I think you can use just algebra,
and no trig.  Think I'll try to do it that way, just for fun,,
since she's already done it the trig way.

Let 

(a + bi)³ = -i, where a, b are real.

a³ + 3a²bi + 3ab²i² + b³i³ = -i

a³ + 3a²bi - 3ab² - b³i = -i

Equating real parts:  | Equating imaginary parts:
                      |
a³ - 3ab² = 0         |  3a²b - b³ = -1
a(a²-3b²) = 0         |
a = 0; a²-3b² = 0     |
           a² = 3b² ->  3(3b²)b - b² = -1
                            9b³ - b³ = -1
                                 8b³ = -1
                                  b³ = -1/8
       a² = 3(1/2)² <-             b = -1/2
       a² = 3/4
        a = ±√3/2

So the three cube roots of -i are:

a + bi

Using a = 0, b = -1/2

a + bi = (-1/2)i

Using a = ±√3/2, 

a + bi = ±√3/2 - (1/2)i

(-1/2)i + [√3/2 - (1/2)i] + [-√3/2 - (1/2)i] =

(-1/2)i + √3/2 - (1/2)i -  √3/2 - (1/2)i = 0

----

Edwin