Real analysis when solving a complex number problem? We are to solve $(x-1)^3+8=0$ in the complex set to do so we bring 8 to the RHS and cube root on both sides $$(x-1)=2(-1)^{1/3}$$
The RHS is of three different forms $(-1,\omega,\omega^2)$
But the LHS is just written in a single form why do we do that ?
I'm nothing close to a mathematician so forgive me for my mathmatical illiteracy .
 A: Let's pick another, less complex example. When solving $x^2 = 1$, we conclude $x = \pm 1$. So, why does the left hand side have a single value, whereas the right has two values?
When you solve an equation like $x^2 = 1$, you are finding all possible values of $x$ that satisfy this equation. You start with the assumption that $x$ is a single value that satisfies the equation $x^2 = 1$. We can then conclude, logically, from this assumption, that $x = 1$ or $x = -1$. It's not that $x$ has both values, it has one value or the other (but not both).
Now, in your example, there is an issue where complex exponentiation can produce multiple answers. In real numbers, we understand that $x^{1/3}$ refers to the one and only cube root of $x$, i.e. the unique number $y \in \Bbb{R}$ such that $y^3 = x$. In the complex numbers, there is no unique $y \in \Bbb{C}$ such that $y^3 = x$ (for $x \neq 0$). As such, you should interpret $2(-1)^{1/3}$ as one of three possible complex numbers, much like you would treat $\pm 1$ as one of two real numbers.
