<s>[INST] An interesting question of algebraic manipulation 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?

 [/INST] Because $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j-3\sum_{1\leq i<j<k\leq n}a_ia_ja_k.$$
We can get it by the following way.
Firstly, it's obvious that $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j+K\sum_{1\leq i<j<k\leq n}a_ia_ja_k$$ for some real $K$.
But for $a_1=a_2=...=a_n=1$ we obtain:
$$n^3=n+3n\cdot\frac{n(n-1)}{2}+K\cdot\frac{n(n-1)(n-2)}{6},$$
which gives $K=-3$.
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