solve the equation in $\Bbb C $ \begin{array}{l}{\text {Solve in } \mathbb{C}}: \\ {x^{2}+\left(\frac{x}{x+1}\right)^{2}=3} \\ {\text { my try: }} \\ {x^{2}(x+1)^{2}+x^{2}=3(x+1)^{2}} \\ {x^{2}\left(x^{2}+2 x+1\right)-3\left(x^{2}+2 x+1\right)+x^{2}=0} \\ ({x^{2}-3 )\left(x^{2}+2 x+1\right)+x^{2}=0} \\ {x^{4}+2 x^{3}-x^{2}-6 x-3=0} \\ {\text { Now what should I do? }}\end{array}

A strange solution that uses symmetry reasoning to transform the problem.
The argument $\frac{x}{1+x}$ stick out and can be set as a new variable. In fact, let's throw in a minus sign for good measure (especially if you recognize it as one of the functions that is inverse of itself).
$$u=-\frac{x}{1+x}\quad \Rightarrow \quad x=-\frac{u}{1+u}$$
If you put this in, you get the same equation for $u$, meaning that both $x$ and $u$ are solutions of this equation. Therefore, instead of one fourth order equation, we have two completely symmetric equations for two variables:
$$\begin{align}x^2+u^2&=3 \\
xu+u+x&=0\quad (\text{from }u=-\frac{x}{1+x})
\end{align}
$$
Adding twice the second equation to the first one gives us
$$(x+u)^2+2(x+u)-3=0$$
which is factorized by hand:
$$\boxed{x+u=\{-3,1\}}$$
Subtracting twice the second equation gives
$$(x-u)^2=3+2(u+x)$$
and using the just derived solutions,
$$\boxed{x-u=\pm\sqrt{\{-3,5\}}}$$
Averaging the two boxed results gives you all four solutions:
$$x=\frac{-3\pm i\sqrt{3}}{2}$$
$$x=\frac{1\pm \sqrt{5}}{2}$$
Note that computing $u$ must give you the same results (that was the premise of this method), and indeed, the $\pm$ in the second boxed equation ensures that swapping the sign just gives you the associated $u$ to each $x$.
