Question 981152
Find the cube roots of 27(cos 279° + i sin 279°).

My Work so far:

(27*(cos 279 + isin279)^1/3= 27^1/3 * (cos 279  + i sin 279) ^ 1/3
27 ^ 1/3 = cube root 27 = 3
3*(cos 279 + i sin 279) ^1/3= 3(cos 279/3 + i sin 279/3)= 3(cos 93 + i sin 93) 

Is my answer right?
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Your answer is right, but it's only 1 of the cube roots.
There are 2 more.