<s>[INST] Expressing $(-8)^{\frac13}$ in polar form I want to express $(-8)^{\frac{1}{3}}$ in polar and cartesian coordinates.
What I did was to solve the equation $-8 = r^3e^{3i\theta}= r^3(\cos(3\theta)+i\sin(3\theta))$ which implies that I must solve the equations
$$
r^3\cos(3\theta) = -8
$$
and the equation
$$
r^3\sin(3\theta)= 0
$$
The latter gives me $\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3},\dotsc$ and so using $\theta = 0$, using the equation $r^3\cos(3\theta) = -8$, I get $r = -2$ and in this case the polar and cartesian form is $-2$, then using $\theta = \frac{\pi}{3}.$ I get $r =2$ and so the polar form is $2e^{\frac{\pi}{3}}$ and the cartesian form is $1+i \sqrt{3}$ and finally using $\theta = \frac{2\pi}{3}$ I get $r=-2$ and in polar form the answer is $-2e^{\frac{2\pi}{3}}$ and in cartesian coordinates the answer is $-1 + i \sqrt{3}$. However, the answers in my book are different mostly by signs, like the first answer is $2$ instead of $-2$. Can anyone explain why I am getting this sign errors?
 [/INST] Your work all seems correct, your book may have a typo. $2$ is definitely NOT a third root of $-8$, as $2^3 = 8$.
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