There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=0, find them.
+655 I think you meant x^6=64, nbot x^6=0... Here you go!
The obvious two solutions to this equation is 2 and -2 since, (±2)6 = 64 but the question asked for the complex numbers too, so there are more solutions.
The first step in order to approach this problem is to factor x6 = 64.
We can re-write this problem into x6 - 26 = 0
And also, (x3)2-(23)2=0
After using difference of cubes and sum of cubes(search it up if you don't know what it is)
I got: (x-2)(x^2+2x+2^2)*{(x+2)(x^2-2x+2^2)}=0
We need to solve ±(x-2)=0,
(x^2+2x+2^2)=0
(x^2-2x+2^2)=0
After you do, you have your six conjugates. I'll leave the solving for you!
64^(1/6) =
+ OR - 2
2 e^((i π)/3)≈1.0000 + 1.7321 i
2 e2 e^(-(2 i π)/3)≈-1.0000 + 1.7321 i
2 e^(-(2 i π)/3)≈-1.0000 - 1.7321 i
2 e^(-(i π)/3)≈1.0000 - 1.7321 i
+118687 Thank you supermanaccz and guest,
Superman, I have never seen it done your way, there are usually several ways to solve the same problem and it is interesting to see alternate routes. 
There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=0, find them.
The 6 points will be equally distributed on the complex plane at an absolute distance of 2 units.
A revolution is 2pi radians so the points will be 2pi/6 = pi/3 radians apart.
Also you know that one of them is \(2 = 2e^0 = 2e^{(0\pi/3)i}\)
so here are your roots.
\(\sqrt[6]{2}=2e^{(0\pi/3)i}, \quad2 e^{1(\pi/3)i}, \quad2 e^{2(\pi/3)i}, \quad2 e^{3(\pi/3)i},\ \quad2 e^{4(\pi/3)i}, \quad2 e^{5(\pi/3)i}\\ \sqrt[6]{2}=2e^{(\pi/3)i}, \quad2 e^{(\pi/3)i}, \quad2 e^{(2\pi/3)i}, \quad2 e^{\pi i},\ \quad2 e^{(4\pi/3)i}, \quad2 e^{(5\pi/3)i}\\\)
I've done it exactly the same way as our guest only he/she has gone from -pi to +pi and I have gone from 0 to 2pi. I am not sure if there is any convention on this, but the answers are the same.
