I think the problem is clear from the title. It included a hint which suggested reducing it to the case $A = 1$, but so far I've come up with little.
It is quite direct from the statement that $A, B, C$ are $n$th roots of unity. I also know that the sum of $n$th roots of unity must be zero, but haven't found a way to link it to 3 dividing $n$.
Any help will be appreciated. Thank you
On
You know that the $n$-roots of unity are forming a $n$-regular polygone on the plane. The condition $A+B+C=0$ is equivalent to say that the points $A,B,C$ are forming an equilateral triangle on the plane.
Now the regular $n$-polygones such that you can find an equilateral triangle with three of its vertices must have $3k$ vertices. (For some $k\in \mathbb{N}_0$)
Perhaps do you look for a less geometric argument ?
On
Let $A=\cos\frac{2\pi m}{n}+i\sin\frac{2\pi m}{n}$, $B=\cos\frac{2\pi k}{n}+i\sin\frac{2\pi k}{n}$ and $C=\cos\frac{2\pi l}{n}+i\sin\frac{2\pi l}{n}$,
where $n\geq m\geq k\geq l\geq0$.
Thus, $$\cos\frac{2\pi m}{n}+\cos\frac{2\pi k}{n}+\cos\frac{2\pi l}{n}=0$$ and $$\sin\frac{2\pi m}{n}+\sin\frac{2\pi k}{n}+\sin\frac{2\pi l}{n}=0.$$ Hence, $$\left(\cos\frac{2\pi m}{n}+\cos\frac{2\pi k}{n}\right)^2+\left(\sin\frac{2\pi m}{n}+\sin\frac{2\pi k}{n}\right)^2=\cos^2\frac{2\pi l}{n}+\sin^2\frac{2\pi l}{n}$$ or $$\cos^2\frac{2\pi m}{n}+2\cos\frac{2\pi m}{n}\cos\frac{2\pi k}{n}+\cos^2\frac{2\pi k}{n}+\sin^2\frac{2\pi m}{n}+2\sin\frac{2\pi m}{n}\sin\frac{2\pi k}{n}+\sin^2\frac{2\pi k}{n}=\cos^2\frac{2\pi l}{n}+\sin^2\frac{2\pi l}{n}$$ or $$2+2\cos\frac{2\pi(m-k)}{n}=1$$ or $$\cos\frac{2\pi(m-k)}{n}=-\frac{1}{2},$$ which gives that $n$ divided by $3$.
After dividing through by $A$ and renaming, we get $B^n=C^n=1$ and $1+B+C=0$, so that $B+C=-1$.
Then if $B=r+si$, $C=(-1-r)-si$, and the fact that $|B| = |C|=1$ quickly shows that $r = -\frac{1}{2}$. Thus $B$ is a cube root of unity.