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$$ \begin{align} a = {7 \choose 0}+{7 \choose 3}+{7 \choose 6}\\ b = {7 \choose 1}+{7 \choose 4}+{7 \choose 7}\\ c = {7 \choose 2}+{7 \choose 5} \end{align} $$ 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.

Asaf Karagila
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user42819
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    You could use the identity $a^3+b^3+c^3-3abc= \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$. However, I feel that if you just simply calculate the values of $a,b,c$ that'll be much easier. All of them are bounded above by $2^7=128$. Hence, it shouldn't be difficult – Anju George May 10 '19 at 15:43
  • Also, as has been pointed out in the answer, $a=b$ and $c=2{7\choose 2}$. Using these facts may reduce the number of computations you have to perform – Anju George May 10 '19 at 15:44
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    @AnjuGeorge Here and in other similar cases of trisecting binomial sums you get $(a-b)^2+(b-c)^2+(c-a)^2=2$ making the right hand side $a+b+c$ (and so $2^7$ in this particular case) , because your three subsums are of the form ${k,k,k\pm1}$ – Henry May 10 '19 at 16:02

3 Answers3

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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 $$2^N=a^3+b^3+c^3-3abc.$$

Angina Seng
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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 $\binom{7}{2}=\binom{7}{5}$.

pwerth
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It is worth noting that

  1. because $\tbinom{n}{k}=\tbinom{n}{n-k}$ we have $a=b$,
  2. 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 $2^7=128$.

Servaes
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