<s>[INST] 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: 
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
 [/INST] As $-1\le x\le1$
WLOG $x=\cos2t,0\le2t\le\pi,\sin2t=\sqrt{1-x^2}$
$$\implies\sqrt{1+\sin2t}[(2\cos^2t)^{3/2}+(2\sin^2t)^{3/2}]=2+\sin2t$$
As $\sin t,\cos t\ge0$ and $(\sin t+\cos t)^2=1+\sin2t$
$$2\sqrt2(\cos t+\sin t)(\cos^3t+\sin^3t)=2+\sin2t$$
$$\sqrt2(1+\sin2t)(2-\sin2t)=2+\sin2t$$ which is on rearrangement,  a Quadratic Equation in $\sin2t(\ge0)$
as $\cos^3t+\sin^3t=(\cos t+\sin t)(\cos^2t+\sin^2t-\sin t\cos t)=\dfrac{(\cos t+\sin t)(2-\sin2t)}2$
</s>