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

Substituting $$a=\sqrt{1+x},b=\sqrt{1-x}$$ and using that $$a^2+b^2=2$$ and $$a^3+b^3=(a+b)(a^2+b^2-ab)$$ we get$$\sqrt{1+ab}(a+b)(2-ab)=2+ab$$
And we get by squaring $$(1+ab)(2+2ab)(2-ab)^2=(2+ab)^2$$
and with $$u=ab$$ we get
$$2(1+u)^3(2-u)^2=(2+u)^2$$
The solutions are $$\left\{\left\{x\to
   -\sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
   \sqrt{97-68 \sqrt{2}}}\right\},\left\{x\to
   \sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
   \sqrt{97-68 \sqrt{2}}}\right\}\right\}$$
