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

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 $c^2=1-x^2$

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 $\sf{x}$.

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 $$\left\{\left\{x\to -\sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4} \sqrt{97-68 \sqrt{2}}}\right\},\left\{x\to \sqrt{-\frac{7}{4}+\frac{3}{\sqrt{2}}-\frac{1}{4} \sqrt{97-68 \sqrt{2}}}\right\}\right\}$$