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

Let $p = A+B = \sqrt{1+x}+\sqrt{1-x}$. We have
$$ p^2 = 2+2\sqrt{1-x^2}$$
$$ p^3 = \sqrt{(1-x)^3}+\sqrt{(1+x)^3} + 3p\sqrt{1-x^2}$$
so 
$$ 1+ \sqrt{1-x^2} = \frac12 p^2$$
$$ \sqrt{(1-x)^3}+\sqrt{(1+x)^3} = -\frac12 p^3 +3p$$
and we get
$$ \frac{p}{\sqrt{2}} (-\frac12 p^3 +3p) = 1 + \frac12 p^2$$
$$ - \frac{1}{2\sqrt{2}}p^4 +\frac{3\sqrt{2}-1}{2} p^2 - 1 = 0$$
which can be solved for $p$.
