Find the value of x such that
$$(1 - x)^{\frac{3}{2}} + (1 + x)^{\frac{3}{2}} + 2(\sqrt{1 - x^2}) = (2 + 2\sqrt{1 - x^2})^{\frac{3}{2}}$$

Let $$\sqrt{\ 1-x}=a\ \&\ \sqrt{\ 1+x}=b$$

R.H.S = $$(2 + 2\sqrt{1 - x^2})^{\frac{3}{2}}$$

=$$(1+x+1-x + 2\sqrt{(1 - x)(1+x)})^{\frac{3}{2}}$$

=$$(a^2+b^2+ 2ab)^{\frac{3}{2}}$$

=$$\left(a+b\right)^{2\times\ \frac{3}{2}}$$ 

=$$\left(a+b\right)^3$$

L.H.S = $$(1 - x)^{\frac{3}{2}} + (1 + x)^{\frac{3}{2}} + 2(\sqrt{1 - x^2})$$

= $$a^3+b^3+2ab$$

=$$\left(a+b\right)^3-3ab(a+b)+2ab$$

Equating L.H.S & R.H.S, we get

$$\left(a+b\right)^3-3ab(a+b)+2ab = \left(a+b\right)^3$$

$$a+b=\frac{2}{3}$$

substituting the values of 'a' & 'b' we get

$$\sqrt{\ 1-x}+\sqrt{\ 1+x}=\frac{2}{3}$$

squaring on both sides, we get

$$1-x+1+x+2\sqrt{\ 1-x^2}=\frac{4}{9}$$

$$2\sqrt{\ 1-x^2}=\frac{4}{9}-2$$

$$\sqrt{\ 1-x^2}=\frac{-7}{9}$$

Squaring on both sides

$$1-x^2=\frac{49}{81}$$

$$x^2=\frac{32}{81}$$

$$x=\frac{4\sqrt{\ 2}}{9}$$

Hence, option (D) is the answer.