Solving $(1-x)^3 = -1$ over the complex field

Question

What are the solutions of: $(1-x)^3 = -1$ over $\mathbb{C}$?

We have one real solution which is $2$ so there are two complex solutions.

Answer 1

Using the fact that $a^3 + 1 = (a+1)(a^2 - a +1)$, we have $$(1-x)^3 + 1 = 0 \iff \Big((1-x)+1\Big)\Big((1-x)^2 - (1-x) + 1\Big) = (2 - x)(x^2 - x + 1)\\ \implies x = 2\, \text{ or }\, x^2 - x + 1 = 0$$

Use the quadratic formula on the remaining quadratic to solve for the remaining two roots.

Answer 2

Using Euler Formula

$$ e^{i(2k+1)\pi}=-1+i0$$

$$(1-z)^3 = -1=e^{i(2k+1)\pi}$$ where $k\in \mathbb Z$

$$\begin{align}(1-z) &= \left(e^{i(2k+1)\pi}\right)^{1/3}\\ &= e^{i\pi/3},e^{3i\pi/3},e^{5i\pi/3}\\ \end{align}$$ So We have

$$\begin{align}z &= 1-e^{3i\pi/3}=2\\ z &= 1-e^{i\pi/3}=\frac {1-i\sqrt{3}}{2}\\ z &= 1-e^{5i\pi/3}=\frac {1+i\sqrt{3}}{2}\\ \end{align}$$