SOLUTION: 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

Notice that in the instructions the word "roots" is plural.
In complex numbers there are always "n nth roots" of any 
complex number except 0. 

As the other tutor pointed out, you have found only one cube root.  
However there are 3 cube roots.  You must begin by adding 360°k 
to the angle since the sines and cosines are the same when we 
add any multiple of 360° to the angle.




must first be written as



Then we do what you did.  We take the real 1/3 power (or 
cube root) of 27 and multiply the angle by 





Now we choose k as any three consecutive integers, say 0,1,and 2.

One cube root will be found by substituting k=0



Cube root #1:    , which was the only one you found.

Use k=1 (which amounts to adding 120° to 93°

The second cube root will be found by substituting k=1
[Using k=1 amounts to adding 120° to 93°]



cube root #2:   ,

Use k=2 (which amounts to adding 120° to 213°

The third and final cube root will be found by substituting k=2
[Using k=2 amounts to adding 120° to 213°]



cube root #3:   ,

If we plot those cube roots on a graph, where the real part
is the x-coordinate and the imaginary part is the y-coordinate,
we find that they are three vectors each 3 units long and equally 
spaced around a circle of radius 3 like spokes of a wheel.  



Edwin