6. If $a^{2}+b^{2}+c^{2}=a b+b c+c a$, what is $(a+2 b-3 c)^{2009}$?
Solution. If we multiply the given equation by 2 and move everything to the left side, we get

$$
2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a=0
$$

If we rearrange the equation, we get

$$
a^{2}+b^{2}-2 a b+b^{2}+c^{2}-2 b c+c^{2}+a^{2}-2 c a=0
$$

or

$$
(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0
$$

from which it follows that $a=b=c$

so the desired value is equal to zero.