Moving from $x^2+x+1=0$ to $x+1+\frac1x=0$ seems to introduce a solution, how is this possible? I recently saw this: ($x\in\mathbb{C}$)

  
*
  
*$$x^2+x+1=0,x\neq0,\pm1$$
  
*$$x+1+\frac{1}{x}=0$$
  Plugging (2) into (1) we get $x^2-\frac{1}{x}=0$ so $x=1$
  

Why does this happen? I know that eq. (1) has solutions $x=\frac{-1\pm\sqrt{-3}}{2}$, and that these still are solutions for eq. (2), but where is the extra solution $x=1$ coming from? (And why is that step not valid? Is it like squaring both sides?)
 A: Another way to look at this.


*

*$$x^2 + x + 1 = 0, x \neq 0, \pm 1$$


*$$x + 1 + \frac{1}{x} = 0$$


*$$x^2 - \frac{1}{x} = 0$$


*$$x = 1$$

We're asking for 'the solutions' to 1. What does that mean? It means, find the set of numbers, $X \subset \mathbb{C}$ with
$$x \in X \iff 1.$$
We know that

*

*$\implies$ 2.

And clearly 2. $\implies$ 1.
So 1. $\iff$ 2.
We know that, together,

*

*and 2. $\implies$ 3.

BUT, it is not the case that


*$\implies$ 1.

This is an assumption that is implicit in the 'proof' given.
We also know that


*$\implies$ 3.

BUT, again, it is not the case that


*$\implies$ 4.

With these implications made explicit, it should be clear why this reasoning is unsound. We are effectively making the claim that 4. $\implies$ 1!
The particular fallacy is called Affirming the Consequent because we 'affirm' the right hand of an implication, and fallaciously deduce from that that the left hand must be true.
This kind of fallacy is particularly common when 'equational' reasoning is used without making the implications explicit.
A: In effect, you have taken the equation
$$x^2+x+1=0,$$
divided by $x$ to get
$$x+1+\frac1x=0$$
and then multiplied by $x-1$ to get
$$x^2-\frac1x=0.$$
The extraneous root $1$ is the solution of $x-1=0$, and $x-1$
is the factor you multiplied by.
A: If $$x^2-\dfrac{1}{x}=0,$$ then $$\frac{x^3-1}{x}=0;$$ but note that $$\frac{x^3-1}{x} =\frac{(x-1)(x^2+x+1)}{x}=0,$$ so you just basically multiplied both sides of $(1)$ by $(x-1)$.
A: When you perform equation (1) - equation (2), you are multiplying $\left(1-\frac1x\right)$ to both sides of equation (1):
$$\begin{align*}
x^2+x+1 &= 0\\
(x^2+x+1)\left(1-\frac1x\right) &= 0\\
x^2 - \frac1x &= 0
\end{align*}$$
So you introduced the root $\frac1x = 1$, i.e. $x=1$.
