Complete the square on the left-hand side

$\begin{aligned} x^2 + \left(\frac x{x-1}\right)^2 & = 8 & \small \color{#3D99F6}{\text{As }a^2+b^2 = (a+b)^2 - 2ab} \\ \left( x + \frac x{x-1}\right)^2 - \frac {2x^2}{x-1} & = 8 \\ \left(\frac {x^2-x+x}{x-1}\right)^2 - \frac {2x^2}{x-1} & = 8 \\ \left(\frac {x^2}{x-1}\right)^2 - 2\left(\frac {x^2}{x-1}\right) & = 8 \\ \left(\frac {x^2}{x-1}\right)^2 - 2\left(\frac {x^2}{x-1}\right) + 1 & = 9 \\ \left(\frac {x^2}{x-1} - 1 \right)^2 & = 3^2 \\ \frac {x^2}{x-1} - 1 & = \pm 3 \\ \frac {x^2}{x-1} & = 1\pm 3 \end{aligned}$

$\implies \begin{cases} x^2 -4x+4 = 0 & \implies x = 2 \\ x^2 +2x-2 = 0 & \implies x = -1 \pm \sqrt 3 \end{cases}$

Therefore, the sum of three roots $=2 - 1+\sqrt 3 -1 -\sqrt 3 = \boxed{0}$

Adding $2x\frac{x}{x-1}$ to both sides, we get:
$x^{2} +2x\frac{x}{x-1} + (\frac{x}{x-1})^{2} = 8 + 2x\frac{x}{x-1}$

$(x+ \frac{x}{x-1})^2 - 2x\frac{x}{x-1} -8 =0$

$(\frac{x^{2}}{x-1})^{2} - 2(\frac{x^{2}}{x-1}) - 8 = 0$

Now we'll substitute: $\frac{x^{2}}{x-1} = t$

And we get a quadratic equation $t^{2} -2t -8 =0$ The solutions to this equation are 4 and -2. Going back and plugging those values in we get

$\frac{x^{2}}{x-1} = 4$ , which transforms into $(x-2)^{2}=0$ , which has one solution: $x_1= 2$ .

And the other case $\frac{x^{2}}{x-1} =- 2$ , which is equivalent to $x^{2} +2x -2=0$ . This equation has two solutions: $x_2= -1 + \sqrt{3}$ and $x_3= -1 - \sqrt{3}$ .

Finally, we have $x_1+x_2+x_3= 2 -1 + \sqrt{3} -1 - \sqrt{3}$

$\boxed{x_1+x_2+x_3 = 0}$

I refuse to complete the square :P

$x^2+\left(\dfrac{x}{x-1}\right)^2 = 8\\ \dfrac{x^2}{x^2-2x+1} = 8-x^2\\ x^2=(8-x^2)(x^2-2x+1)\\ x^2=8x^2-x^4-16x+2x^3+8-x^2\\ x^4-2x^3-6x^2+16x-8=0$

From the Rational Root theorem, we find that $x=2$ is a solution. Factorize it out:

$x^4-2x^3-6x^2+12x+4x-8=0\\ x^3(x-2)-6x(x-2)+4(x-2)=0\\ (x-2)(x^3-6x+4)=0$

Again, from the Rational Root theorem, we find that $x=2$ is a repeated solution. Factorize it out again:

$(x-2)(x^3-2x^2+2x^2-4x-2x+4)=0\\ (x-2)\left(x^2(x-2)+2x(x-2)-2(x-2)\right)=0\\ (x-2)(x-2)(x^2+2x-2)=0$

$x=2,\;x=\dfrac{-2\pm\sqrt{2^2-4(1)(-2)}}{2(1)} = \dfrac{-2\pm\sqrt{12}}{2} = -1 \pm \sqrt{3}$

There are $3$ distinct solutions, and their sum is $2-1+\sqrt{3}-1-\sqrt{3}=\boxed{0}$

First, we observe that 1 can't be a solution. We multiply then the equation by $(x-1)^2$ .

$x^2(x-1)^2 + x^2 = 8(x-1)^2$

Second, we recall that the sum of the roots of a polynomial with leading coefficient 1 is equal to the opposite of the second-to-leading coefficient. Here, the polynomial is of order 4 and we easily see the coefficient of $x^3$ is -2. Then the sum of the 4 roots is 2.

Third, we know there are only 3 distinct solutions. 1 is not a root (easy check) so we know the polynomial problem has a double root. Observe then that if $x$ is a solution, then $\frac{x}{x-1}$ is one too. Also, applying the transformation twice, we get back to $x$ . Then the number of roots must be even except if one of the roots is stable by the transformation. Obvious stable points are 0 and 2. 0 is not a solution of the initial problem while 2 is. Hence 2 must be the double root. The sum of the three distinct solutions is therefore $2 -2 =0$ .