Solve $x^2+x^2t^2+2xt^2+t^2=1$ for $x$. Solving
$$x^2+x^2t^2+2xt^2+t^2=1$$
for $x$ yields
$$x=\frac{1-t^2}{t^2+1}.$$
I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.
What is the method?

This equation can be written as  $x^2+t^2(x+1)^2=1$ which $\implies t^2=\frac{(1+x)(1-x)}{(1+x)^2}\implies t^2=\frac{(1-x)}{(1+x)}$ (assuming $x\neq -1$) $ \implies t^2(1+x)=1-x\implies t^2+t^2x=1-x\implies x(t^2+1)=1-t^2\implies x=\frac{1-t^2}{1+t^2}$
Here, we have to check whether $x=-1$ satisfies the equation separately. Putting in the equation $x=-1$, it satisfies the equation. Thus  $x=-1, \frac{1-t^2}{1+t^2}$  
