Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found: 
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$ 
Therefore, it is legal to divide each term by $x$: 
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks. 

Transformations you apply to an equation may introduce alien solutions.
Taking an extreme example,
$$x=0\implies 0=0$$ which is satisfied by all $x$ !
So you may apply transformations, but validate the solutions using the original equation.

In your example, you establish
$$x^2+x+1=0\implies x^3-1=0.$$
But as $$x^3-1=0=(x-1)(x^2+x+1),$$ nothing is wrong if you ignore the first factor.
