Find the cube roots of 27(cos 279° + i sin 279°).

The solution would be like this for this specific problem:

27(cos 279° + i sin 279°).
(cosx +i sinx) = cox(nx)) + i sin(nx)
(27×(cos 279+i sin 279))1/3=27 1/3×(cos 279+i sin 279)1/3
2713=27−−√3=?
3×(cos279+i sin279)1/3
3×(cos 279+i sin 279)13=3(cos 279/3+i sin 279/3)

279/3 = 93
cube root = 3(cos 93 + i sin 93)

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21.4 i did it already for s1

step-by-step explanation: