The value of $$\left(\frac{1+\sin\frac{2\pi}{9}+i\cos\frac{2\pi}{9}}{1+\sin\frac{2\pi}{9}-i\cos\frac{2\pi}{9}}\right)^3$$ is:

1. Transformation to Polar Form

The given expression is:

$$\left( \frac{1+\sin \frac{2\pi}{9}+i \cos \frac{2\pi}{9}}{1+\sin \frac{2\pi}{9}-i \cos \frac{2\pi}{9}} \right)^3$$

We use the co-function identities $$\sin\theta\ =\ \cos\left(\frac{\pi}{2}-\theta\ \right)\ $$ and  $$\cos\theta\ \ =\ \sin\left(\frac{\pi}{2}-\theta\ \right)$$ to convert the terms:

$$\sin\frac{2\pi}{9}\ =\ \cos\left(\frac{\pi}{2}-\frac{2\pi}{9}\right)\ =\ \cos\ \frac{5\pi}{18}$$

$$\cos\frac{2\pi}{9}\ =\ \sin\left(\frac{\pi}{2}-\frac{2\pi}{9}\right)\ =\ \sin\ \frac{5\pi}{18}$$

Let $$\theta = \frac{5\pi}{18}$$. The expression inside the parentheses becomes:

$$\frac{1 + \cos \theta + i \sin \theta}{1 + \cos \theta - i \sin \theta}$$

2. Conversion to $$e^{i\theta}$$ using Half-Angles

We apply the half-angle identities:$$1+\cos\theta\ \ \ \ =\ 2\cos^2\ \frac{\theta}{2}\ $$  and $$\sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$$.

For the Numerator:

$$1+\cos\theta\ \ +\ i\sin\theta\ \ =\ 2\cos^2\ \frac{\theta}{2}\ +i\left(2\sin\ \frac{\theta}{2}\cos\ \frac{\theta}{2}\ \right)$$

$$= 2\cos\frac{\theta}{2} \left( \cos\frac{\theta}{2} + i \sin\frac{\theta}{2} \right) = 2\cos\frac{\theta}{2} \cdot e^{i\theta/2}$$

For the Denominator:

$$1+\cos\theta\ \ -\ i\sin\theta\ \ =\ 2\cos^2\ \frac{\theta}{2}\ -i\left(2\sin\ \frac{\theta}{2}\cos\ \frac{\theta}{2}\ \right)$$

$$= 2\cos\frac{\theta}{2} \left( \cos\frac{\theta}{2} - i \sin\frac{\theta}{2} \right) = 2\cos\frac{\theta}{2} \cdot e^{-i\theta/2}$$

3. Simplification and Power Application

Dividing the numerator by the denominator:

$$\frac{2\cos\frac{\theta}{2} \cdot e^{i\theta/2}}{2\cos\frac{\theta}{2} \cdot e^{-i\theta/2}} = \frac{e^{i\theta/2}}{e^{-i\theta/2}} = e^{i(\frac{\theta}{2} - (-\frac{\theta}{2}))} = e^{i\theta}$$

Now, raise this result to the power of $3$:

$$(e^{i\theta})^3 = e^{i3\theta}$$

4. Final Evaluation

Substitute $$\theta = \frac{5\pi}{18}$$ back into the exponential form:

$$e^{i3\left(\frac{5\pi}{18}\right)} = e^{i\frac{5\pi}{6}}$$

Convert the result back to rectangular form:

$$e^{i\frac{5\pi}{6}} = \cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6}$$

Using the trigonometric values for $$\frac{5\pi}{6}$$ (which is $$150^\circ$$):

$$\cos\ \frac{5\pi}{6}\ =\ -\frac{\sqrt{3}}{2},\ \ \sin\ \frac{5\pi}{6}\ =\ \frac{1}{2}$$

$$e^{\frac{i5\pi}{6}}=\ -\frac{\sqrt{3}}{2}+\ \frac{i}{2}$$

Factor out $$-\frac{1}{2}$$ to match the format of Option C:

$$= -\frac{1}{2}(\sqrt{3} - i)$$