<s>[INST] 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.
 [/INST] There is no need to rewrite the expression - you can simply plug in the values of $a,b,c$ directly. Since 
$$\binom{n}{k}=\frac{n!}{k!(n-k)!},$$
we get (I'll leave the algebra to you)
$$a=1 + 35 + 7 = 43$$
$$b=7 + 35 + 1 = 43$$
$$c=21 + 21 =42$$
so that $a^{3}+b^{3}+c^{3}-3abc= 128$. Note that the useful identity 
$$\binom{n}{k}=\binom{n}{n-k}$$
can be used to reduce the number of computations above e.g. $\binom{7}{3}=\binom{7}{4}$ and $\binom{7}{2}=\binom{7}{5}$.
</s>