### Instruction: Find the cubic polynomial, in $x,$ with integer coefficients that has $\cos 20^\circ$ as a root.  The coefficient of $x^3$ should be positive, and the coefficients should have no common factor other than 1.### Response: To solve this problem, I will use the fact that $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ for any angle $\theta$.
If I let $\theta = 20^\circ$, then I get $\cos 60^\circ = 4\cos^3 20^\circ - 3\cos 20^\circ$.
But I know that $\cos 60^\circ = \frac{1}{2}$, so I can substitute that and rearrange the equation to get $4\cos^3 20^\circ - 3\cos 20^\circ - \frac{1}{2} = 0$.
Now, I want to find a cubic polynomial in $x$ that has $\cos 20^\circ$ as a root, so I can replace $\cos 20^\circ$ by $x$ and multiply both sides by 2 to get rid of the fraction.
This gives me $8x^3 - 6x - 1 = 0$ as the desired polynomial.
I can check that this polynomial has integer coefficients, a positive leading coefficient, and no common factor other than 1, as required by the problem.
Therefore, this is the final answer.