Find the cube roots of 8i. Write the answer in a + bi form.

a0 =  ∛8 [ cos [ (pi/2)/3] +  i *sin [(pi/2)/3 ]  = ∛8 [ cos [ (pi/6)] +  i *sin [(pi/6) ] =

2 [ √3/2 + 1/2 * i ] =    √3  + 1i

a1  = ∛8 [ cos ( (pi/2)/3 + 2pi/3 )  + i sin ( (pi/2 )/3 + 2pi/3 ) ]   =

2 [ cos (5pi/6)  + i sin (5pi/6) ] =  2 [ - √3/2  +  (1/2)*i ]  =   - √3  +1i

a2  = ∛8 [ cos ( (pi/2)/3 + 4pi/3 )  + i sin ( (pi/2 )/3 + 4pi/3 ) ] =

2 [ cos(3pi/2) + i sin (3pi/2) ] =   2 [ -1 i ]  = -2i

like this

It's B

Step-by-step explanation:

Just took the quiz

step by step solution :

step   1   :

1.1     4 = 22

(4)3 = (22)3 = 26

equation at the end of step   1   :

  ((23) • 26)2

step   2   :

equation at the end of step   2   :

  (23 • 26)2

step   3   :

final result :

  218

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