Problem in polynomials related to permutation of number of solution The zeros of $$f(x) = {x^6+2x^5+3x^4+5x^3+8x^2+13x+21} $$ are distinct complex numbers. Compute the average value of A+BC+DEF over all possible permutations (A,B,C,D,E,F) of these six numbers.
Please help me solve this problem in a simple way as I am only a high school student.
 A: There is a way to solve this using Vieta's formulas. Let $e_i$ be the elementary symmetric polynomials of the roots, which are up to a sign the coefficients of the polynomial by Vieta's formulas. The sum in question is
$$120e_1+48e_2+36e_3$$
and the average is this divided by $6!=720$. The $120$ comes from the fact that the value of $A$ will repeat $120$ times due to the $5!$ permutations of the other values, the particular $BC$ will occur $48$ times due to the fact that we can exchange $B$ and $C$ and that there are $4!=24$ permutations of the other values, and there are $3!=6$ permutations stabilizing $DEF$, times the $3!=6$ permutations of the other values.
By the formulas, $e_1=-2$, $e_2=3$, and $e_3=-5$, so the sum is $-240+144-180=-276$ and the average is $-\frac{23}{60}$.
For a proof of Vieta's formula for the three coefficients we need, note the polynomial is
$$(x-A)(x-B)(x-C)(x-D)(x-E)(x-F)$$
Multiplying this out (which I won't do in full), we get
$$x^6-(A+B+C+D+E+F)x^5+(AB+AC+AD+\cdots+EF)x^4-(ABC+ABD+\cdots +DEF)x^3+\cdots+ABCDEF$$
The point is that the coefficient of $x^5$ is negative the sum of the roots, the coefficient of $x^4$ is the sum of all products of distinct pairs of roots, and the coefficient of $x^3$ is negative the sum of all products of three distinct roots. With my notation, this is
$$x^6-e_1x^5+e_2x^4-e_3x^3+\cdots+e_6$$
