<s>[INST] Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
 $$a^3 -(b^3)i + 3(a^2)bi -3(b^2)a$$
Equating real and imaginary parts, we get:
$$  a^3 -3(b^2)a = 1  \tag{1}$$
$$  b^3 -3(a^2)b = 1  \tag{2}$$ 
Factoring $(1)$, $(2)$ to get some roots in the process:
$$a(a^2 - 3(b^2)) = 1 \tag{1'}$$
$$b(b^2 - 3(a^2)) = 1 \tag{2'}$$
From $(1')$, the roots are possibly given by equations below:
$$a = 1                      \tag{3}$$
$$  - \text{or} -  $$ // 'or' is used in logical or sense, i.e. either or both
$$ a^2 - 3(b^2) = 1          \tag{4}$$
Substituting $a = 1$ in $(4)$, we get:
$b = 0$; which cannot be a possible solution, as the right side of 
question has $b = -1$. Hence, $a \ne 1$ also.
Next, trying for the possible value from $(2')$.
$$ b = 1                      \tag{5} $$
$$  -\text{or}-  $$ // 'or' is used in logical or sense, i.e. either or both
$$ b^2 - 3(a^2) = 1          \tag{6} $$
Substituting $b = 1$ in $(6)$, we get:
$$ a = 0; $$
which cannot be a possible solution, as the right side of question 
has $a = 1$. Hence, $b \ne 1$ also.
I am unable to solve it further, as no solution emerges from the 
two equations - $(1')$, $(2')$.
 [/INST] Hint:
$$a^3-3b^2a=b^3-3a^2b=1 \implies a^3-b^3+3a^2b-3b^2a=0$$ which implies 
$$(a-b)(a^2+ab+b^2)+3ab(a-b)=0 \implies (a-b)(a^2+4ab+b^2)=0.$$
As such, $a=b$ or $a^2 + 4ab+b^2 = a^2+4ab+4b^2-3b^2 = 0 \implies a = (-2 \pm \sqrt{3})b.$
If $a=b$ then $a^3-3b^2a = -2a^3=1 \implies \color{red}{a=b=-2^{-1/3}}$.
I leave the calculations for $a=(-2 \pm \sqrt{3})b$ to you.
An alternative way of computing the remaining roots (as also suggested by some other people) is given below.
Let $x_0 = -2^{-1/3}-2^{-1/3}i=-2^{-1/3}(1+i)$. We just found out that $x_0^3 = 1-i$. Now
$$x^3 = 1-i \implies \left(\frac{x}{x_0}\right)^3 = \frac{1-i}{x_0^3}=1.$$
Let $y = x/x_0$. We are looking for the solution of $y^3-1=0$. But 
$$y^3-1 = (y-1)(y^2+y+1)=0 \implies y=1 \quad \text{ or }\quad y^2+y+1=0. $$
Now $y=1 \implies x=x_0$, and $$y^2+y+1 = 0 \implies y = \frac{x}{x_0}= \frac{-1\pm \sqrt{3}i}{2}.$$
As such, the remaining cube roots of $1-i$ are $$-2^{-1/3}(1+i)\left(\frac{-1\pm \sqrt{3}i}{2}\right).$$
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