1. **State the problem:** Prove that $$\arccos\sqrt{\frac{2}{3}} - \arccos\frac{\sqrt{6}+1}{2\sqrt{3}} = \frac{\pi}{6}$$. 2. **Recall the formula:** For any angles $A$ and $B$, the difference of arccosines can be related using the cosine difference formula: $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ If we let $$A = \arccos\sqrt{\frac{2}{3}}, \quad B = \arccos\frac{\sqrt{6}+1}{2\sqrt{3}},$$ then $$A - B = \frac{\pi}{6} \iff \cos(A - B) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}.$$ 3. **Evaluate $\cos A$ and $\cos B$:** $$\cos A = \sqrt{\frac{2}{3}}, \quad \cos B = \frac{\sqrt{6}+1}{2\sqrt{3}}.$$ 4. **Find $\sin A$ and $\sin B$ using $\sin \theta = \sqrt{1 - \cos^2 \theta}$:** $$\sin A = \sqrt{1 - \left(\sqrt{\frac{2}{3}}\right)^2} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}.$$ $$\sin B = \sqrt{1 - \left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)^2}.$$ Calculate inside the square root: $$\left(\frac{\sqrt{6}+1}{2\sqrt{3}}\right)^2 = \frac{(\sqrt{6}+1)^2}{4 \cdot 3} = \frac{6 + 2\sqrt{6} + 1}{12} = \frac{7 + 2\sqrt{6}}{12}.$$ So, $$\sin B = \sqrt{1 - \frac{7 + 2\sqrt{6}}{12}} = \sqrt{\frac{12 - 7 - 2\sqrt{6}}{12}} = \sqrt{\frac{5 - 2\sqrt{6}}{12}}.$$ 5. **Simplify $\sin B$:** Note that $$5 - 2\sqrt{6} = (\sqrt{6} - 1)^2,$$ because $$(\sqrt{6} - 1)^2 = 6 - 2\sqrt{6} + 1 = 7 - 2\sqrt{6},$$ which is close but not equal. So instead, keep as is for now. 6. **Calculate $\cos(A - B)$:** $$\cos(A - B) = \cos A \cos B + \sin A \sin B = \sqrt{\frac{2}{3}} \cdot \frac{\sqrt{6}+1}{2\sqrt{3}} + \frac{1}{\sqrt{3}} \cdot \sqrt{\frac{5 - 2\sqrt{6}}{12}}.$$ 7. **Simplify the first term:** $$\sqrt{\frac{2}{3}} \cdot \frac{\sqrt{6}+1}{2\sqrt{3}} = \frac{\sqrt{2/3} (\sqrt{6}+1)}{2\sqrt{3}} = \frac{(\sqrt{2}/\sqrt{3})(\sqrt{6}+1)}{2\sqrt{3}} = \frac{\sqrt{2}(\sqrt{6}+1)}{2 \cdot 3} = \frac{\sqrt{2}(\sqrt{6}+1)}{6}.$$ 8. **Simplify the second term:** $$\frac{1}{\sqrt{3}} \cdot \sqrt{\frac{5 - 2\sqrt{6}}{12}} = \sqrt{\frac{5 - 2\sqrt{6}}{12 \cdot 3}} = \sqrt{\frac{5 - 2\sqrt{6}}{36}} = \frac{\sqrt{5 - 2\sqrt{6}}}{6}.$$ 9. **Sum the two terms:** $$\cos(A - B) = \frac{\sqrt{2}(\sqrt{6}+1)}{6} + \frac{\sqrt{5 - 2\sqrt{6}}}{6} = \frac{\sqrt{2}(\sqrt{6}+1) + \sqrt{5 - 2\sqrt{6}}}{6}.$$ 10. **Simplify $\sqrt{5 - 2\sqrt{6}}$:** Try to express as $\sqrt{a} - \sqrt{b}$: $$(\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6},$$ so $$\sqrt{5 - 2\sqrt{6}} = \sqrt{3} - \sqrt{2}.$$ 11. **Substitute back:** $$\cos(A - B) = \frac{\sqrt{2}(\sqrt{6}+1) + (\sqrt{3} - \sqrt{2})}{6} = \frac{\sqrt{2}\sqrt{6} + \sqrt{2} + \sqrt{3} - \sqrt{2}}{6} = \frac{\sqrt{12} + \sqrt{3}}{6}.$$ 12. **Simplify $\sqrt{12}$:** $$\sqrt{12} = 2\sqrt{3},$$ so $$\cos(A - B) = \frac{2\sqrt{3} + \sqrt{3}}{6} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}.$$ 13. **Conclusion:** Since $$\cos(A - B) = \frac{\sqrt{3}}{2} = \cos\frac{\pi}{6},$$ we have $$A - B = \frac{\pi}{6}$$ as required. **Final answer:** $$\arccos\sqrt{\frac{2}{3}} - \arccos\frac{\sqrt{6}+1}{2\sqrt{3}} = \frac{\pi}{6}.$$