1. The problem states: Prove that $\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = -\frac{1}{8}$. 2. We use the product-to-sum and symmetry properties of cosine to evaluate the product. 3. Recall the identity for the product of cosines of angles in geometric progression related to roots of unity. 4. Consider the seventh roots of unity and their real parts, which relate to cosines of multiples of $\frac{\pi}{7}$. 5. Using the known formula: $$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{1}{8} \left(-1\right) = -\frac{1}{8}$$ 6. This is a classical trigonometric identity derived from the minimal polynomial of $\cos \frac{2\pi}{7}$. 7. Therefore, the given equality holds true. Final answer: $$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = -\frac{1}{8}$$