Mistake in raising power Find roots of:
$$x^{6}\ -\ \left(x-1\right)^{6}=0 \tag {1}$$
I know this equation has $4$ complex roots and exactly one real roots of value $0.5$.
However, my first instinct was to do this:
$$x^{6}\ =\ \left(x-1\right)^{6} \tag{2}$$
"raise both sides to 6-th power" to get:
$$x=x-1\tag{3}$$
Which has no real solution. I see that this wrong. How to avoid this error? Thanks.
Inspired by watching this youtube video
Edit:
I am not asking about how to solve the problem. I want to know
what I did wrong from an Algebraic stand-point. Maybe raising to the power? What is wrong with that?

If you had started with $x^2 - (x-1)^2=0$ then you can expand it to $x^2-x^2+2x-1=0$ and so $x=\frac12$.
But let's take a version of your approach. You might say $a^2=b^2$ clearly implies $a=b$ or $a=-b$; let's write this as $a=\omega b$ where $\omega^2=1$.  We expect there to be be two possible values for $\omega$, namely $+1$ and $-1$.  But if $a=x$ and $b=x-1$ you get $x=\omega(x-1)$ and so $x=  \frac{-\omega}{1-\omega}= 1 - \frac{1}{1-\omega}$; when $\omega=-1$ it gives $x=\frac12$ as expected, while if $\omega=1$ it tries to give $1- \frac{1}{0}$ which is not a finite number, reflecting the fact there is no finite solution to $x=x-1$.
Translate this to your $6$th powers in $(2)$:

*

*$a^6=b^6$ has solutions of the form $a=\omega b$ where $\omega^6=1$;

*there are potentially six such $\omega$ (some complex) of which one is $\omega=1$

*If $a=x$ and $b=x-1$ you get $x=\omega(x-1)$ and again $x=  1 - \frac{1}{1-\omega}$

*So each of the six $\omega$ will give you a different solution for $x$, except $\omega=1$ which tries to give $1-\frac{1}{0}$ (still not a finite number).

