<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] Since $1-i= \sqrt 2 (\cos315^\circ + i\sin315^\circ),$ the cube roots of $1-i$ must be
$$
2^{1/6} (\cos(105^\circ+n120^\circ) + i\sin(105^\circ+n120^\circ))
$$
where the only values of $n$ we need to consider are $0,$ $1,$ and $2.$
If we can believe the tables on this page, we have
$$
\cos105^\circ = -\frac 1 4 (\sqrt 6 - \sqrt 2\,) \quad \text{and} \quad \sin(105^\circ) = \frac 1 4 (\sqrt6 + \sqrt 2\,).
$$
(The page gives $\cos15^\circ = \dfrac 1 4 ( \sqrt 6 + \sqrt 2\,)$ and $\sin15^\circ = \dfrac 1 4 (\sqrt 6 - \sqrt 2\,).$ If you consider $\dfrac 1 2 = \sin30^\circ = 2\sin15^\circ\cos15^\circ,$ then this appears to make sense. And then $\cos105^\circ = -\sin15^\circ$ and $\sin105^\circ = \cos15^\circ$ since $105 = 15 +90.$)
So we should have
$$
\sqrt 2\,\left( -\frac 1 4(\sqrt6-\sqrt2\,) + i \frac 1 4( \sqrt6+\sqrt 2\,) \right)^3 = 1-i.
$$
</s>