number theory - Is this (deceptive) pattern just a coincidence?

I've come across this nice pattern start:
$$
\begin{align}
1^2-2^2+3^2 &= 6\\
1^4-2^4+3^4 &= 66\\
1^6-2^6+3^6 &= 666\\
\end{align}
$$
I read somewhere the third line, I remembered that I already knew the first, I immediately tested the second and I thought "eureka, there is a pattern!". I was already hoping to be able to find an explanation when I realized that the next would-be term fails to follow the "rule" ($1^8-2^8+3^8=6306$). Now, lacking a pattern, I shouldn't need a proof. Nevertheless I find it hard to believe that those three lines can be mere coincidence. I'm wondering whether some sort of "reason" can anyway be found (for them being what they are, and for the next ones not following).

Thank you!

Answer

At the moment it seems that this is an "assisted coincidence". Thanks to Thomas Andrews I know that the expressed quantities are congruent to $6$ mod $60$, and this reduces significantly the possible results, so that numbers like $66$ and $666$ no longer seem unbelievably lucky.

Thanks to Greg Martin it is easy to understand why the pattern couldn't last forever. I'll write the would-be general term of the pattern
$$
1^{2n}-2^{2n}+3^{2n} \stackrel{?}{=} 6\cdot\sum_{k=0}^{n-1}10^k
$$
Dividing by $10^n$ one can see that the right-hand side tends towards $2/3$:
$$

\lim_{n\rightarrow +\infty}\frac{6}{10^n}\cdot\sum_{k=0}^{n-1}10^k = 6\cdot\lim_{n\rightarrow +\infty}\sum_{k=0}^{n-1}\left(\frac{1}{10}\right)^{n-k}= 6\cdot\lim_{n\rightarrow +\infty}\sum_{d=1}^{n}\left(\frac{1}{10}\right)^d=6\cdot\frac{1}{9}=\frac{2}{3}
$$
while the left-hand side tends towards $0$ ( $\sim(3^2/10)^n$ ), having a chance to be equal to the right side only for little $n$'s. I've made a plot to show this, switching to $x$, where one can see the fortunate interceptions at $(1,0.6)$, $(2,0.66)$ and $(3,0.666)$, while after those the curves separate definitively. See images below.

I love Peter's remark that $6$, $66$ and $666$ are triangular numbers, while $6666$ is not, but I don't see a link to the original question.