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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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2 Answers

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
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!