### 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 find a cubic polynomial with $\cos 20^\circ$ as a root, I need to use some trigonometric identities and algebraic manipulations.
One identity that might be useful is the triple angle formula for cosine: $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta.$
If I let $\theta = 20^\circ$, then I get $\cos 60^\circ = 4\cos^3 20^\circ - 3\cos 20^\circ.$
I know that $\cos 60^\circ = \frac{1}{2}$, so I can write $\frac{1}{2} = 4\cos^3 20^\circ - 3\cos 20^\circ.$
I can multiply both sides by 8 to get rid of the fraction: $4 = 32\cos^3 20^\circ - 24\cos 20^\circ.$
I can rearrange this equation to get a cubic polynomial in $\cos 20^\circ$: $32\cos^3 20^\circ - 24\cos 20^\circ - 4 = 0.$
I can divide both sides by 4 to simplify the coefficients: $8\cos^3 20^\circ - 6\cos 20^\circ - 1 = 0.$
Now, I have a cubic polynomial in $\cos 20^\circ$ with integer coefficients and no common factor other than 1.
But I want a cubic polynomial in $x$ with the same properties.
How can I do that?
I can make a substitution: let $x = \cos 20^\circ.$
Then the polynomial becomes $8x^3 - 6x - 1 = 0.$
This is the cubic polynomial I'm looking for.
It has $\cos 20^\circ$ as a root, it has integer coefficients with no common factor other than 1, and it has a positive coefficient of $x^3.$