<s>[INST] What are the three cube roots of $1-i$? I am trying to find the three cube roots of $1-i$ and I am unsure of how to proceed. My math teacher recommended I use de Moivre's formula, however I do not know how to set it up. How do you suggest I continue?
de Moivre's formula
$$(\cos x + i\sin x)^n = \cos(nx) + i\sin(nx)$$
 [/INST] Another way without using DeMoivre's formula that works for cube roots is set
$x=1$ and $y=-i$ and then find any real roots of the following equation
$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$
This expands to
$$a^9 - \frac{3}{4}a^6 - \frac{21}{32}a^3 - \frac{1}{64} = 0$$
and has three real roots  $a_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16}); $$a_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})$; $a_3=\frac{1}{2}(-\sqrt[6]{16})$. 
  Next solve the equations 
$$a_1^3+3a_1b_1^2 = 1$$ 
$$a_2^3+3a_2b_2^2 = 1$$ 
$$a_3^3+3a_3b_3^2 = 1$$ 
(where $1$ is the value set as $x$) for $b_1$, $b_2$ and $b_3$, or $b_1=\pm\frac{1}{4}(\sqrt[6]{432}-\sqrt[6]{16})i$; $b_2=\pm\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i; $$b_3=\pm\frac{1}{2}(\sqrt[6]{16})i$ . So the three cube roots of $$\sqrt[3]{1-i}$$ are
$$a_1+b_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})i$$
$$a_2+b_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i$$
$$a_3+b_3=\frac{1}{2}(-\sqrt[6]{16})+\frac{1}{2}(-\sqrt[6]{16})i$$
Combinations of $a_1$, $a_2$ and $a_3$ and $b_1$, $b_2$ and $b_3$ will also give the three roots of $\sqrt[3]{1+i}$.
</s>