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

Great substitution technique. Notice that $\sf{A=\sqrt{1+x}}$ and $\sf{B=\sqrt{1-x}}$ leads to $$\sf{(A^3+B^3)\sqrt{1+AB}}=2+AB$$ and since $\sf{A^3+B^3=(A+B)(A^2-AB+B^2)}$ and $\sf{A^2+B^2=2\implies (A+B)^2=2(1+AB)}$, we get $$\sf{(A+B)\sqrt{1+AB}=\frac{2+AB}{2-AB}}\implies (1+AB)\sqrt2=1+\frac{2AB}{2-AB}$$ which can be solved for $\sf{AB}$, and thus $\sf{x}$.
