<s>[INST] If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.

In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.

I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to build the equation up from there.
What I got was that $$\cos3A+\cos3B-\cos(3A+3B)=1$$
Expanding, I got $$\cos3A+\cos3B-\cos3A\cos3B-\sin3A\sin3B=1$$
Now, it's possible that I'd be able to factor it somehow by rewriting it all in terms of cosine and arrive at the answer, but is there a better way to solve the problem? Thanks!
 [/INST] We have $$2\cos\frac{3A+3B}{2}\cos\frac{3A-3B}{2}-2\sin^2\frac{3C}{2}=0$$ or
$$\cos\left(\frac{3\pi}{2}-\frac{3C}{2}\right)\cos\frac{3A-3B}{2}-\sin^2\frac{3C}{2}=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}-\sin\frac{3C}{2}\right)=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}+\cos\frac{3A+3B}{2}\right)=0$$ or
$$\sin\frac{3C}{2}\sin\frac{3C}{2}\sin\frac{3C}{2}=0$$ or
$$\prod_{cyc}\left(3\sin\frac{A}{2}-4\sin^3\frac{A}{2}\right)=0$$ or
$$\prod_{cyc}\left(3-4\sin^2\frac{A}{2}\right)=0$$ or
$$\prod_{cyc}\left(3-2(1-\cos{A})\right)=0$$ or
$$\prod_{cyc}\left(1+\frac{b^2+c^2-a^2}{bc}\right)=0$$ or
$$\prod_{cyc}(b^2+c^2+ab-a^2)=0.$$
Can you end now?
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