Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$


My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(1+AB)(A^{3} + B^{3})^{2} = (AB)^{2} + 4AB + 4 $$
$$ A^{6} + B^{6} + BA^{7}  + A B^{7} = -2 (AB)^{4} - 2(AB)^{3} + (AB)^{2} + 4AB + 4 $$
I also have tried using$A = (1+x), B = (1-x)$ , and some others, but none solves the problem.


I am now trying $A = (1+x)$ and $(1-x) = -(1+x) + 2 = 2 - A$, so: 
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$

Note: $1-x^2\ge 0 \Rightarrow -1\le x\le 1$.
Square both sides and denote $1-x^2=t^2, -1\le t\le 1$:
$$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} \Rightarrow \\
(1+\sqrt{1-x^2})(2+6x^2+2\sqrt{(1-x^2)^3}=5-x^2+4\sqrt{1-x^2} \Rightarrow \\
(1+t)(2+6(1-t^2)+2t^3)=4+t^2+4t \Rightarrow \\
2t^4-4t^3-7t^2+4t+4=0 \Rightarrow \\
t\approx 0.93, x\approx \pm 0.38.$$
Note: Other roots are rejected as they are outside of domain.
