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 $\sum\limits_{i=1}^n a_i^3 = 3C_n$.
Let
- $g(x) = x^5 f\left(\frac1x\right) = x^5+5x^4-2x^3+3x^2-4x+1$.
- $S = \{ a,b,c,d,e \}$ be the roots of $f(x)$.
- $T = \{ \frac1a, \frac1b, \frac1c, \frac1d, \frac1d \}$ be the roots of $g(x)$.
- For $I \subset S$ and $J \subset T$, let $\lambda_I = \prod_{\lambda \in I}\lambda$ and $\mu_J = \prod_{\mu \in J}\mu$.
The polynomial we seek equals to
$\quad\displaystyle\;F(x) \stackrel{def}{=} \prod_{I \subset S,|I| = 3}(x - \lambda_I)$.
Define a similar polynomial for $g$,
$\quad\displaystyle\;G(x) \stackrel{def}{=} \prod_{J \subset T,|J| = 2}(x - \mu_J)$.
By Vieta's formula, we have $abcde = -1$, this implies
$$F(x) = \prod_{I\subset S,|I|=3} \left(x + \frac{1}{\lambda_{S \setminus I}}\right) = \prod_{J\subset T,|J|=2}(x + \mu_J) = G(-x)$$
The problem comes down to given $g(x)$, how to compute $G(x)$ whose roots are
product of distinct pairs of roots of $g(x)$.
It will be hard to relate the coefficients of $g$ and $G$ directly. However, there is a simple relation between the power sums. More precisely, for any
$k \in \mathbb{Z}_{+}$, let
- $P_k(g) \stackrel{def}{=} \sum_{\mu \in T} \mu^k$ be the sum of roots of $f(x)$ raised to power $k$.
- $P_k(G) \stackrel{def}{=} \sum_{J \subset T,|J|=2} \mu_J^k$ be the sum of roots of $G(x)$ raised to power $k$.
We have
$$P_k(G) = \frac12( P_k(g)^2 - P_{2k}(g))\tag{*1}$$
To make following descriptions more generic, let $n = 5$ and $m = \frac{n(n-1)}{2}$.
Define coefficients $\alpha_k, \beta_k$ as follow:
$$g(x) = x^n - \sum\limits_{k=1}^n \alpha_k x^{n-k}
\quad\text{ and }\quad
G(x) = x^m - \sum\limits_{k=1}^m \beta_k x^{m-k}$$
Following are the steps to compute coefficients $\beta_k$ from coefficients $\alpha_k$ manually.
- Compute $P_k(g)$ using
Newton's identity for $1 \le k \le 2m$.
$$P_k(g) = \sum_{j=1}^{\min(n,k-1)} \alpha_j P_{k-j}(g) + \begin{cases}
k \alpha_k, & k \le n\\
0, & \text{otherwise}\end{cases}
$$
Compute $P_k(G)$ from $P_k(g)$ using $(*1)$.
Compute $\beta_k$ from $P_k(G)$ using Newton's identities again:
$$\beta_k = \frac1k\left( P_k(G) - \sum_{j=1}^{k-1} \beta_j P_{k-j}(G) \right)$$
I am lazy, I implement above logic in maxima (the CAS I use) and compute these numbers. The end result is
$$F(x) = x^{10}-2x^9+19x^8-112x^7+82x^6+97x^5-15x^4+58x^3+3x^2+3x+1$$
If one has access to a CAS, there is a quicker way to get the result.
For example, in maxima, one can compute the resultant between $g(t)$ and $g\left(-\frac{x}{t}\right)$ using the command resultant(g(t), g(-x/t), t)).
The resultant of two polynomials is essentially the GCD of them over the polynomial ring. It vanishes when and only when the two polynomials share a root. When the resultant between $g(t)$ and $g\left(-\frac{x}{t}\right)$ vanishes, $x$ either equals to $-\mu^2$ for a root $\mu \in T$ or $-\mu\nu$ for some $\mu, \nu$ in $T$.
If one ask maxima to factor output of above command, the result is
$$-(x^5+29x^4-34x^3+3x^2+10x+1)F(x)^2$$
The first factor is nothing but $\prod\limits_{\mu \in T}(x + \mu^2)$, this confirm the expression we get for $F(x)$ is the product $\prod\limits_{J \subset T,|J| = 2}(x + \mu_J)$ we desired.
Best Answer
A nifty way to get started; you want to compute the coefficients of a polynomial of degre $10$. The coefficient of $x^9$ is the negative of the sum of the roots, and by Vieta's formulae we can immediately read off $$abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=2,$$ from the original polynomial, so the coefficient of $x^9$ equals $-2$.
Similarly, the constant term is the product of all the roots, which is $$(abc)(abd)(abe)(acd)(ace)(ade)(bcd)(bce)(bde)(cde)=(abcde)^6=(-1)^6=1,$$ so the constant term equals $1$.
For the remaining steps, identities of the form $$(abc)(cde)=(abcde)c=-c,$$ go a long way in simplifying the computations; this shows that the coefficient of $x^8$ equals $-4$, and the coefficient of $x$ equals $-3$. Still a bit tedious, but no more than a few minutes worth of work.