Find the 3 cube roots of 8

[facto2]

Hello, Would you please check my work to verify it is correct. Ive attached a document with the problem. The problem is: Find the 3 cube roots of 8. I used De Moivre's theorem. My instructor marked me down and only commented that the K's are OK but the square root of 8 is 2. I know that the square root of 8=2 but did i do it right using this method? thank you very much for your time and assistance

 

[Deleted member 4993]

Hello, Would you please check my work to verify it is correct. Ive attached a document with the problem. The problem is: Find the 3 cube roots of 8. I used De Moivre's theorem. My instructor marked me down and only commented that the K's are OK but the square root of 8 is 2. I know that the square root of 8=2 -- since when? one of the CUBE roots of 8 is 2 but did i do it right using this method? thank you very much for your time and assistance

Find three cube roots of 8

\(\displaystyle 8^{\frac{1}{3}} \,\)

\(\displaystyle = \, \left [8\cdot \left [1\right ]\right]^{\frac{1}{3}}\)

\(\displaystyle = \, \left [8\cdot \left [-cos(\pi) - i \cdot sin(\pi)\right ]\right]^{\frac{1}{3}}\)

\(\displaystyle = \, 2\cdot \left [e^{i\cdot\pi\cdot (2n-1)}\right]^{\frac{1}{3}}\)

So the three solutions are

\(\displaystyle 2\cdot e^{-i\cdot{\frac{\pi}{3}}\)

\(\displaystyle 2\cdot e^{-i\cdot{\frac{2\pi}{3}}\)

\(\displaystyle 2\cdot e^{-i\cdot{\frac{3\pi}{3}}\)