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 us first simplify the expression. Assume $a^2=1-x$ $ $ & $ $ $b^2=1+x$
Please note that $a^2+b^2=2$
Now our equation becomes $\sqrt{1+ab}(a^3+b^3)=2+ab$
$\sqrt{\frac{a^2+b^2+2ab}{2}}(a+b)(a^2+b^2-ab)=2+ab$  which further simplifies to
$\frac{(a+b)^2}{\sqrt{2}}(2-ab)=2+ab$
$\frac{(2+2ab)(2-ab)}{\sqrt{2}}=2+ab$
Find $ab$ and use the equation $a^2b^2=1-x^2$
Hope this will be helpful!
