I am trying to solve the quadratic equation $x^2 + x + 1 = 0$. $x^2 = -1 - x $ $\iff x = -\frac{1}{x} - 1$, assuming $x\neq 0$.
Substituting that into the original equation gives $x^2 + (-\frac{1}{x} -1) + 1 = 0$
$x=1$ is a solution to this second equation, but it isn't a solution to the first. How did this extra solution arise? Which step wasn't reversible and why wasn't it?
The source of this extraneous solution is the same as that of those encountered while completing the square: mistaking $ P \implies Q $ to mean the converse $ Q \implies P $, where $P$ is $ x^2 + x + 1 = 0 $ and $Q$ is $ x^2 + (-\frac{1}{x} - 1) + 1 = 0 $.
Try going backwards from $Q$ to $P$ yourself: you can't, at least not without the proposition $R$ $ x = -\frac{1}{x} - 1 $ (which is rightly derivable from $P$, but not from $Q$).
A more rigorous way to go about solving for $x$ would be to assume $ P \iff Q \ \& \ R $. So now you don't have to go back to $P$ to check if your solution is extraneous, but you do have the added restriction of $R$ which you have to keep in mind.
Since $R$ eliminates $ x = 0 $, and there are no other solutions for $Q$; you are forced to conclude that $P$ does not have any solutions either, and indeed your quadratic equation has no solutions in $\mathbb{R}$ (the two roots are complex: $-\sqrt[3]{-1} = -i^{\frac{2}{3}} $ and $ (-1)^\left(2/3\right) = i^{\frac{4}{3}}$).
That would mean $x = undef$ (or rather $\not\exists x \in \mathbb{R}$), I'm not sure what what the mathematical meaning or validity of $(x−1)(x^2+x+1)=0(x−1)$ would be other than just be saying $undef = undef$. (Could someone formalize what's happening here?)
Also, if we're trying to go backward from $x^2 + \frac{1}{x} = 0$ (which does have the solution 1), we can indeed multiply both sides by the identity $\frac{(x - y)(x^2 + xy + y^2}{(x^3 - y^3)}$ as it is equal to $1$; but we must ensure that the denominator we're dividing by is not 0 (as if we break the rules, we're kicked out of the nice world of our number theory), which in the case of $x = 1$ is indeed that exactly.
You have $x=-x^2-1$ and $x=-\frac1x-1$,
Let $A=x$, $B=-x^2-1$ and $C=-\frac1x-1$. You start with $A=B$ and $A=C$. You conclude that $B=C$. The trouble is going backwards from $B=C$ to $A=B$ and $A=C$.
Your reasoning is:
$\hspace{57pt} x^{2}+x+1\overset{1}{=}0 \ \left(x\in\mathbb{R}\right)$
$\overset{1}{\iff} \hspace{30pt} \text{Rearrange:} \ x^{2}\overset{2}{=}-x-1$
$\overset{2}{\iff} \hspace{30pt} \text{Since }x=0\text{ doesn't solve } \overset{1}{=} \text{, we have } x\neq0. \text{ So, divide } \overset{2}{=} \text{ by } x \text{ to get}$:
$$ x\overset{3}{=}-1-\frac{1}{x}$$
$\overset{3}{\iff} \hspace{30pt} \text{Now plug } \overset{3}{=} \text{ into } \overset{1}{=} \text{to get:}$
$$x^{2}+\left(-1-\frac{1}{x} \right)+1\overset{4}{=}0.$$
And now we see that $x=1$ solves $\overset{4}{=}$ but not $\overset{1}{=}$. Why?
The error in the above chain of reasoning is in $\overset{3}{\iff}$, which should instead be $\overset{3}{\implies}$:
And so, it is at $\overset{3}{\iff}$ that extraneous solution(s) may be (and indeed is) introduced.