Those stubborn roots

$\begin{aligned} \frac {x+\sqrt 3}{\sqrt x + \sqrt{x+\sqrt 3}} + \frac {x-\sqrt 3}{\sqrt x + \sqrt{x-\sqrt 3}} & = \sqrt x & \small \color{#3D99F6} \text{Let }a = x + \sqrt 3, \ b = x - \sqrt 3 \\ \frac a{\sqrt x + \sqrt a} + \frac b{\sqrt x - \sqrt b} & = \sqrt x \\ \frac {a(\sqrt x - \sqrt a)}{(\sqrt x + \sqrt a)(\sqrt x - \sqrt a)} + \frac {b(\sqrt x + \sqrt b)}{(\sqrt x - \sqrt b)(\sqrt x + \sqrt b)} & = \sqrt x \\ \frac {a(\sqrt x - \sqrt a)}{x-a} + \frac {b(\sqrt x + \sqrt b)}{x-b} & = \sqrt x & \small \color{#3D99F6} \text{Note that }x- a = - \sqrt 3, \ x-b = \sqrt 3 \\ \frac {a(\sqrt x - \sqrt a)}{-\sqrt 3} + \frac {b(\sqrt x + \sqrt b)}{\sqrt 3} & = \sqrt x \\ (b-a)\sqrt x + a\sqrt a + b\sqrt b & = \sqrt {3x} & \small \color{#3D99F6} \text{Note that }b- a = - 2\sqrt 3 \\ -2\sqrt{3x} + a\sqrt a + b\sqrt b & = \sqrt {3x} \\ a\sqrt a + b\sqrt b & = 3 \sqrt {3x} & \small \color{#3D99F6} \text{Squaring both sides} \\ a^3 + b^3 + 2ab\sqrt{ab} & = 27x & \small \color{#3D99F6} \text{Putting back }a = x + \sqrt 3, \ b = x - \sqrt 3 \\ (x+\sqrt 3)^3 + (x-\sqrt 3)^3 + 2(x^2-3)^\frac 32 & = 27x \\ 2x^3 + 18x + 2\sqrt{x^6-9x^4+27x^2-27} & = 27x \\ 2\sqrt{x^6-9x^4+27x^2-27} & = 9x - 2x^3 & \small \color{#3D99F6} \text{Squaring both sides again} \\ 4x^6-36x^4+108x^2-108 & = 4x^6 - 36x^4 - 81x^2 \\ 27x^2 & = 108 \\ \implies x & = \boxed{2} & \small \color{#3D99F6} \text{Note that }x \ge 0 \text{ for } \sqrt{x} \text{ to be defined.} \end{aligned}$