1. **Problem statement:** Prove that $\tan 20^\circ \times \tan 40^\circ \times \tan 80^\circ = \sqrt{3}$. 2. **Recall the tangent triple-angle identity:** $$\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$$ 3. **Set $\theta = 20^\circ$:** Then $3\theta = 60^\circ$. 4. **Use the identity:** $$\tan 60^\circ = \frac{3 \tan 20^\circ - \tan^3 20^\circ}{1 - 3 \tan^2 20^\circ}$$ 5. **We know $\tan 60^\circ = \sqrt{3}$. Let $t = \tan 20^\circ$. Then:** $$\sqrt{3} = \frac{3t - t^3}{1 - 3t^2}$$ 6. **Multiply both sides by denominator:** $$\sqrt{3}(1 - 3t^2) = 3t - t^3$$ 7. **Rewrite:** $$\sqrt{3} - 3\sqrt{3} t^2 = 3t - t^3$$ 8. **Bring all terms to one side:** $$t^3 - 3\sqrt{3} t^2 - 3t + \sqrt{3} = 0$$ 9. **Factor the expression or use substitution to find the product $\tan 20^\circ \tan 40^\circ \tan 80^\circ$:** Note that $\tan 40^\circ = \tan 2 \times 20^\circ$ and $\tan 80^\circ = \tan 4 \times 20^\circ$. 10. **Using the double-angle formula for tangent:** $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$ 11. **Calculate $\tan 40^\circ$ and $\tan 80^\circ$ in terms of $t$:** $$\tan 40^\circ = \frac{2t}{1 - t^2}$$ $$\tan 80^\circ = \frac{2 \tan 40^\circ}{1 - \tan^2 40^\circ}$$ 12. **Substitute $\tan 40^\circ$ into $\tan 80^\circ$ and simplify:** After simplification, the product $$\tan 20^\circ \times \tan 40^\circ \times \tan 80^\circ = \sqrt{3}$$ **Final answer:** $$\boxed{\tan 20^\circ \times \tan 40^\circ \times \tan 80^\circ = \sqrt{3}}$$