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: 
$$\sqrt{1 + \sqrt{A(2-A)}}\left(\sqrt{(A)^{3}} + \sqrt{(2-A)^{3}} \right) = 2 + \sqrt{A(2-A)} $$
 A: First, this is stated as an equation to solve (for $x$) rather than an identity to be shown.
So with $a=\sqrt {1+x}$ and $b=\sqrt {1-x}$ we have $$a^2+b^2=2$$ and $$(a+b)^2=a^2+2ab+b^2=2(1+ab)$$ and $$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(2-ab)$$
Then $$\sqrt {1+ab}\cdot (a^3+b^3)=\frac {\sqrt 2}2(a+b)(a+b)(2-ab)=\sqrt 2(1+ab)(2-ab)$$
If we then put $c=ab$ the equation to solve is then $$\sqrt 2(1+c)(2-c)=2+c$$ which is a straightforward quadratic in $c$. Then solve for $x$ by noting $c^2=1-x^2$
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 $\sf{x}$.
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 $$\left\{\left\{x\to
   -\sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
   \sqrt{97-68 \sqrt{2}}}\right\},\left\{x\to
   \sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4}
   \sqrt{97-68 \sqrt{2}}}\right\}\right\}$$
A: 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 $\cos^3t+\sin^3t=(\cos t+\sin t)(\cos^2t+\sin^2t-\sin t\cos t)=\dfrac{(\cos t+\sin t)(2-\sin2t)}2$
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!
A: Note: $1-x^2\ge 0 \Rightarrow -1\le x\le 1$.
Square both sides and denote $1-x^2=t^2, -1\le t\le 1$:
$$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} \Rightarrow \\
(1+\sqrt{1-x^2})(2+6x^2+2\sqrt{(1-x^2)^3}=5-x^2+4\sqrt{1-x^2} \Rightarrow \\
(1+t)(2+6(1-t^2)+2t^3)=4+t^2+4t \Rightarrow \\
2t^4-4t^3-7t^2+4t+4=0 \Rightarrow \\
t\approx 0.93, x\approx \pm 0.38.$$
Note: Other roots are rejected as they are outside of domain.
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 $p$.
