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.

It is worth noting that


*

*because $\tbinom{n}{k}=\tbinom{n}{n-k}$ we have $a=b$,

*because $\sum_{k=0}^n\tbinom{n}{k}=2^n$ we have $c=2^7-a-b=2^7-2a$.


It follows that
\begin{eqnarray*}
a^3+b^3+c^3-3abc
&=&2a^3+(2^7-2a)^3-3a^2(2^7-2a)\\
&=&(2-2^3+3\cdot2)a^3+(3\cdot2^9-3\cdot2^7)a^2-3\cdot2^{15}a+2^{21}\\
&=&2^73^2a^2-3\cdot2^{15}a+2^{21}\\
&=&2^7((3a)^2-2^8(3a)+2^{14})\\
&=&2^7(3a-2^7)^2.
\end{eqnarray*}
A quick computation shows that $3a-2^7=1$ and so the result is $2^7=128$. 
