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?

 A: 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:

If $A_n = 0$, then $\sum\limits_{i=1}^n a_i^3 = 3C_n$

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 $\sum\limits_{i=1}^n a_i^3 = 3C_n$.
A: Instead of claiming it is "well known", actually prove it.
$0^3 = (a+b+c)^3 =$
$(a^3 + b^3 + c^3)+ 3(a^2b + a^2c + b^2a + b^2 c + c^2a + c^2b) + 6abc=$
$(a^3 + b^2 + c^3) +3((a^2b + a^2c + abc) + (b^2a+b^2c + bac) + (c^2a+c^2b + cab))- 3abc=$
$(a^3 + b^3 + c^2) + 3(a(ab+ac+bc) + b(ba + bc + ac) + c(ac+bc + ab))-3abc=$
$(a^3 + b^3 + c^2) + 3(a+b+c)(ab+ac+bc)-3abc=$
$(a^3 + b^3 + c^2) + 3*0*(ab+ac+bc)-3abc=(a^3 + b^3 + c^2) -3abc$ so $a^3 + b^3 + c^2 = -3abc$
It's much the same with multiple values but it's a variable tracking night mare:
$(a + b + c+d + e)^3 = $
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3(a^2b + a^2c + a^2d + a^2e + b^2a+b^2c+b^2d +b^2e + c^2a+c^2b + c^2d + c^2e + d^2a+d^2b + d^2c+d^2e+e^2a+e^2b +e^2c +e^2d) + 6(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=$
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3((a^2b + a^2c + a^2d + a^2e + abc + abd+abe+acd+ace + ade) + (b^2a+b^2c+b^2d +b^2e+bac + bad+bae+bcd+bce+bde) + (c^2a+c^2b + c^2d + c^2e+cab+cad+cae+cbd+cbe+cde) + (d^2a+d^2b + d^2c+d^2e+dab + dac+dae+dbc+dbe+dce)+(e^2a+e^2b +e^2c +e^2d+eab+eac+ead + ebc+ebd+ecd)) - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=$
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3(a+b+c+d+e)(ab+ac+ad+ae+bc+bd+be+cd+ce) - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=0$
$(a^3 + b^3 + c^3 + d^3 + e^3)  - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)$
Doing it with multiple variables is much the same (but extremely easy to get lost in the notation):
$(\sum a_i)^3 = (\sum a_i^3) + 3(\sum_{j\ne k} a_j^2a_k) + 6(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3) + 3(\sum_{\lnot(i=j=k)}a_ia_ja_k)- 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3) + 3[\sum a_i](\sum_{j\ne k} a_ja_k) - 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3)  - 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=0$
A: 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 $K=-3$.
