<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] Someone has pointed out that one solution of $x^3 - (1-i) =0$ is $\dfrac{-1-i}{\sqrt[3]2}.$
Therefore $x+\dfrac{1+i}{\sqrt[3]2}$ is a factor of $x^3 - (1-i).$ So we can do long division:
$$
\begin{array}{cccccccccc}
& & & & x^2 & - & \dfrac{1+i}{\sqrt[3]2}x & + & i\sqrt[3]2  \\ \\
x + \dfrac{1+i}{\sqrt[3]2} & \Big) & x^3 & + & 0x^2 & + & 0x & - & (1-i) \\
& & x^3 & + & \dfrac{1+i}{\sqrt[3]2} x^2 \\  \\  \\
& & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & & & - & (1-i) \\
& & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & - & i x \sqrt[3]2 \\  \\  \\
& & & & & & ix\sqrt[3]2 & - & (1-i) \\
& & & & & & ix\sqrt[3]2 & - & (1-i) \\  \\  \\
& & & & & & & & 0 \\
\hline
\end{array}
$$
Thus we have found that
$$
x^3-(1-i) = \left( x + \frac{1+i}{\sqrt[3]2} \right) \left( x^2 - \frac{1+i}{\sqrt[3]2} x + i\sqrt[3]2 \right).
$$
The other two solutions are now solutions of a quadratic equation. If we write the equation as $ax^2+bx+c=0$ then we have $a=1,$ $b= - \dfrac{1+i}{\sqrt[3]2},$ and $c=i\sqrt[3]2,$ and
$$
b^2-4ac = -3i\sqrt[3]2.
$$
So what is $\pm\sqrt{b^2-4ac~} \text{ ?}$ We just saw that $(1+i)^2 = 2i,$ i.e. $\pm\sqrt{2i} = \pm(1+i),$ so that gives us $\pm\sqrt{b^2-4ac~}.$
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