### Instruction:
If $\arccos x + \arccos 2x + \arccos 3x = \pi,$ then $x$ satisfies a cubic polynomial of the form
\[ax^3 + bx^2 + cx + d = 0,\]where $a,$ $b,$ $c,$ and $d$ are integers, and $a \neq 0.$  Find the smallest possible value of $|a| + |b| + |c| + |d|.$

### Response:

Let's use the identity $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$. That's a good idea. We can apply it to $\arccos x + \arccos 2x$. We have $\cos(\arccos x \pm \arccos 2x) = \cos(\arccos x)\cos(\arccos 2x) \mp \sin(\arccos x)\sin(\arccos 2x)$. So $\cos(\arccos x) = x$. What are $\cos(\arccos 2x)$ and $\sin(\arccos 2x)$? Since $\arccos 2x$ is the angle whose cosine is $2x$, $\cos(\arccos 2x) = 2x$. How about $\sin(\arccos 2x)$? We have $\sin^2(\arccos 2x) + \cos^2(\arccos 2x) = 1$. So $\sin(\arccos 2x) = \pm \sqrt{1 - \cos^2(\arccos 2x)} = \pm \sqrt{1 - 4x^2}$. Right. Now we need to pick the right sign. Since $\arccos 2x$ is between $0$ and $\pi$, $\sin(\arccos 2x) \geq 0$. So $\sin(\arccos 2x) = \sqrt{1 - 4x^2}$. Now we can put all the pieces together. We have $\cos(\arccos x \pm \arccos 2x) = \cos(\arccos x)\cos(\arccos 2x) \mp \sin(\arccos x)\sin(\arccos 2x) = x*2x \mp \sqrt{1-x^2}\sqrt{1-4x^2}$. That's correct. Now we need to pick the right sign. We have $\arccos x + \arccos 2x + \arccos 3x = \pi$. So $\arccos x + \arccos 2x = \pi - \arccos 3x$. Good. So $\cos(\arccos x + \arccos 2x) = \cos(\pi - \arccos 3x)$. And $\cos(\pi - \theta) = -\cos(\theta)$. So we have $\cos(\arccos x + \arccos 2x) = -\cos(\arccos 3x)$ Now we can use the identity we found earlier. We have $\cos(\arccos x + \arccos 2x) = -\cos(\arccos 3x) = -3x$. Let's do that. So we have $-3x = x*2x - \sqrt{1-x^2}\sqrt{1-4x^2}$. Right. So $x*2x + 3x = \sqrt{1-x^2}\sqrt{1-4x^2}$. That's correct. Now we square both sides. We get $(2x^2 + 3x)^2 = (1-x^2)(1-4x^2)$. And that simplifies to $4x^4 + 12x^3 + 9x^2 = 1 - 5x^2 + 4x^4$. We get $4x^4 + 12x^3 + 9x^2 = 1-5x^2+4x^4$. So $12x^3 + 14x^2 - 1 = 0$. Now we just need to find $|a| + |b| + |c| + |d|$. $a$ is the coefficient of $x^3$, which is 12. $b$ is the coefficient of $x^2$, which is 14. $c$ is the coefficient of $x$, which is 0. $d$ is the constant term, which is -1. So $|a| + |b| + |c| + |d| = 12 + 14 + 0 + 1 = 27$. That's the final answer.

