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Find the complex cube roots of 8(cos(4pi/5) + isin(4pi/5)

I just need help. If someone could give me the answer as well as explain it, that would be amazing

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Howard L. | Experienced Math Tutor Specializing in Test Prep & Grade Up SkillsExperienced Math Tutor Specializing in T...
5.0 5.0 (20 lesson ratings) (20)
If z is a complex number, written in polar form as
z=8(cos(4pi/5) + isin(4pi/5)  
then z^(1/3) = (8(cos(4pi/5) + isin(4pi/5))^(1/3)
                   = 8^(1/3)(cos((1/3)(4pi/5+2pi*k)) + isin((1/3)(4pi/5+2pi*k))) where k=0,1,2
when K = 0, arg z= 4pi/15      the root a0 = 2(cos(4pi/15)+isin(4pi/15)) = 1.33826 + 1.48629i
         k = 1, arg z= 14pi/15                a1 =  2(cos(14pi/15)+isin(14pi/15))  = -1.95623 + 0.41582i
         k = 2, arg z= 24pi/15                a2 =  2(cos(24pi/15)+isin(24pi/15))  =  0.61803 - 1.90211i
Polina O. | Math tutor who believes that everybody can be great in math!Math tutor who believes that everybody c...
4.6 4.6 (15 lesson ratings) (15)
DeMoivre's Theorem tells that there are always three cube roots of a complex number, spaced evenly around a circle. 
if w=r(cosΘ+isinΘ), then zk=cube root(r)•(cos((Θ+2πk)/3)+isin((Θ+2πk)/3)), k=0,1,2.
Please google "roots of complex numbers" to find more information on the topic with examples and illustrations.
Good luck!