Suppose that $a=\binom70+\binom73+\binom76,b=\binom71+\binom74+\binom77,c=\binom72+\binom75$. How to algebraically compute $a^3+b^3+c^3-3abc$? 
$$
\begin{align}
a = {7 \choose 0}+{7 \choose 3}+{7 \choose 6}\\
b = {7 \choose 1}+{7 \choose 4}+{7 \choose 7}\\
c = {7 \choose 2}+{7 \choose 5}
\end{align} 
$$
  then $a^3+b^3+c^3-3abc$ is equal to _____.

I tried to write $a^3+b^3+c^3-3abc$ in terms of $a+b+c$ and failed.
$$
\begin{align}
a^3+b^3+c^3-3abc & = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\
& = (2^7)((a+b+c)^2-3(ab+bc+ca))\\
& = (2^7)((2^7)^2-3(ab+bc+ca))
\end{align}
$$
I think the expression should be written in terms of another binomial series which I can not think of.

More generally, let $N\in \Bbb N$ and let
$$a=\sum_{0\le 3k\le N}\binom N{3k},$$
$$b=\sum_{0\le 3k+1\le N}\binom N{3k+1},$$
$$c=\sum_{0\le 3k+2\le N}\binom N{3k+2}.$$
(Yours is the case $N=7$.)
Let $\omega=\exp(2\pi i/3)=\frac12(-1+i\sqrt3)$. Then $\omega^3=1$.
Also, by the binomial theorem
$$a+b+c=(1+1)^N=\sum_{j=0}^N\binom Nj=2^N,$$
$$a+b\omega+c\omega^2=\sum_{j=0}^N\binom Nj\omega^j=(1+\omega)^N,$$
$$a+b\omega^2+c\omega=\sum_{j=0}^N\binom Nj\omega^{2j}=(1+\omega^2)^N.$$
Then
\begin{align}
a^3+b^3+c^3-3abc&=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)\\
&=[2(1+\omega)(1+\omega^2)]^N=2^N
\end{align}
since $(1+\omega)(1+\omega^2)=1$.
ADDED IN EDIT
Here is essentially the same argument, but avoiding complex numbers.
Let $A=\pmatrix{0&1&0\\0&0&1\\1&0&0}$. Then $A^2=\pmatrix{0&0&1\\1&0&0\\0&1&0}$ and $A^3=I$ etc. Then
$$(I+A)^N=\sum_{j=0}^N\binom Nj A^j=\pmatrix{a&b&c\\c&a&b\\b&c&a}.$$
As $\det(I+A)=2$, then taking determinants gives
$$2^N=a^3+b^3+c^3-3abc.$$
