Question 137
where is the error? - Be $x$ $x^{2}+x+1=0$ solution. So $x \neq 0$ and so we can divide both members of the equation by $x$, obtaining $x+1+\frac{1}{x}=0$. From the equation we have that $x+1=-x^{2}$, then $-x^{2}+\frac{1}{x}=0$, that is: $x^{2}=1 / x$ or $x^{3}=1$ and $x=1$. Replacing $x=1$ in the equation $x^{2}+x+1=0$ we find $3=0$ !!!! where do we go wrong?