Equating Radicals

Let $y=\sqrt[3]{1+\sqrt{x}}$ . Thus, $\sqrt{1+\sqrt{2+\sqrt{x}}} = \sqrt[3]{1+\sqrt{x}} \implies \sqrt{1+\sqrt{1+y^3}} = y$ .

This implies $y^2-1=\sqrt{1+y^3}$ , so $y^4 - 2y^2 + 1 = 1+ y^3$ . This simplifies to $y^4 - y^3 - 2y^2 = 0$ , which simplifies to $y^2 (y+1)(y-2) = 0$ .

Therefore, $y=0, -1, 2$ . Testing, we find only $y=2$ works, which implies $x=49$ is the only solution.

Therefore, the sum of all real $x$ satisfying $\sqrt{1+\sqrt{2+\sqrt{x}}} = \sqrt[3]{1+\sqrt{x}}$ is $\boxed{49}$ .