If $a+b+c+d+e=0$ then prove that $a^3+b^3+c^3+d^3+e^3=3(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde)$ Extend this argument to n integers such that if $a_1+a_2+a_3+\cdots a_n=0$ then $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
My try: At first I tried it for the base case. $a+b+c=0$ then it is a well-known fact that $a^3+b^3+c^3=3abc$, which proves the base case.Let us assume that the given statement is true for some $n$. $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
$$a_1^3+a_2^3+a_3^3\cdots a_n^3+a_{n+1}^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
Subtract the two equations we get,
$$a_{n+1}^3=3\left[(a_1a_2a_{n+1}+a_1a_3a_{n+1}+\cdots a_1a_na_{n+1})+(a_2a_3a_{n+1}+a_2a_4a_{n+1}+\cdots a_2a_na_{n+1})+\cdots +(a_{n-1}a_na_{n+1})\right]$$
I am stuck here. I cannot proceed further. Can please anybody help me?
Best Answer
For any $1 \le \ell \le n$, let $Q_\ell = A_\ell(A_\ell^2 - 3 B_\ell) + 3C_\ell$ where $$ A_\ell = \sum_{i=1}^\ell a_i,\quad B_\ell = \sum_{1\le i < j \le \ell} a_i a_j,\quad\text{ and }\quad C_\ell = \sum_{1\le i < j < k \le \ell} a_ia_ja_k $$
What we want to show is equivalent to the statement:
Notice for any $1 < \ell \le n$, we have
$$A_\ell = A_{\ell-1} + a_\ell,\quad B_\ell = B_{\ell-1} + a_\ell A_{\ell-1}\quad\text{ and }\quad C_\ell = C_{\ell-1} + a_\ell B_{\ell-1}$$ This leads to $$\begin{align} C_\ell - A_\ell B_\ell &= (C_{\ell-1} + a_\ell B_{\ell-1}) - (A_{\ell-1} + a_\ell)(B_{\ell-1} + a_\ell A_{\ell-1}) \\ &= C_{\ell-1} - A_{\ell-1} B_{\ell-1} - a_\ell A_{\ell_1} (A_{\ell-1} + a_\ell) \end{align} $$ Notice $$A_\ell^3 = (A_{\ell-1} + a_\ell)^3 = A_{\ell-1}^3 + a_\ell^3 + 3a_\ell A_{\ell_1}(A_{\ell-1} + a_\ell)$$
Multiply $1^{st}$ equation by $3$ and add to $2^{nd}$ equation, we obtain $Q_\ell = Q_{\ell-1} + a_\ell^3$.
Together with the fact $Q_1 = a_1^3$, we have
$$\sum_{i=1}^n a_i^3 = a_1^3 + \sum_{i=2}^n a_i^3 = Q_1 + \sum_{i=2}^n (Q_i - Q_{i-1}) = Q_n = A_n(A_n^2 - 3B_n) + 3C_n$$ When $A_n = 0$, this reduces to the desired identity $\sum\limits_{i=1}^n a_i^3 = 3C_n$.