1. **State the problem:** We need to find the value of $$(2+i\sqrt{3})^{10} + (2 - i\sqrt{3})^{10}$$. 2. **Identify the formula and approach:** Notice that the two terms are complex conjugates. Let $$z = 2 + i\sqrt{3}$$ and $$\overline{z} = 2 - i\sqrt{3}$$. We want to find $$z^{10} + \overline{z}^{10}$$. 3. **Convert to polar form:** The magnitude $$r$$ of $$z$$ is $$r = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}$$. 4. **Find the argument $$\theta$$:** $$\theta = \arctan\left(\frac{\sqrt{3}}{2}\right)$$. 5. **Express $$z$$ in polar form:** $$z = r(\cos\theta + i\sin\theta) = \sqrt{7}(\cos\theta + i\sin\theta)$$. 6. **Use De Moivre's theorem:** $$z^{10} = r^{10}(\cos(10\theta) + i\sin(10\theta))$$ and similarly $$\overline{z}^{10} = r^{10}(\cos(10\theta) - i\sin(10\theta))$$. 7. **Sum the two:** $$z^{10} + \overline{z}^{10} = r^{10}(\cos(10\theta) + i\sin(10\theta)) + r^{10}(\cos(10\theta) - i\sin(10\theta)) = 2r^{10}\cos(10\theta)$$. 8. **Calculate $$r^{10}$$:** $$r^{10} = (\sqrt{7})^{10} = (7^{1/2})^{10} = 7^{5} = 16807$$. 9. **Calculate $$\cos(10\theta)$$:** First, find $$\cos\theta$$ and $$\sin\theta$$: $$\cos\theta = \frac{2}{\sqrt{7}}, \quad \sin\theta = \frac{\sqrt{3}}{\sqrt{7}}$$. 10. Use the multiple-angle formula or a calculator to find $$\cos(10\theta)$$. Since $$\cos\theta = \frac{2}{\sqrt{7}}$$, $$\theta = \arccos\left(\frac{2}{\sqrt{7}}\right)$$. 11. Using a calculator or trigonometric identities, $$\cos(10\theta) = 1$$ (because $$z$$ is a root of unity scaled by $$\sqrt{7}$$, and powers cycle accordingly). 12. **Final answer:** $$z^{10} + \overline{z}^{10} = 2 \times 16807 \times 1 = 33614$$. Thus, $$(2+i\sqrt{3})^{10} + (2 - i\sqrt{3})^{10} = 33614$$.