Bashing unavailable - part 5

I always wanted to solve this systematically. Perhaps the following helps. I used Newton's sums.

Let $P_1 = a + b + c, \quad P_2 \ a^2+b^2+c^2, \quad P_3=a^3+b^3+c^3...$

and $S_1 = a+b+c,\quad S_2 = ab+bc+ca \quad$ and $\quad S_3 = abc$ .

Then we have:

$P_1 = S_1$

$P_2 = S_1P_1-2S_2$

$P_3 = S_1P_2-S_2P_1+3S_3$

$P_4 = S_1P_3-S_2P_2+S_3P_1$

$P_5 = S_1P_4-S_2P_3+S_3P_2$

$P_6 = S_1P_5-S_2P_4+S_3P_3$

$P_7 = S_1P_6-S_2P_5+S_3P_4$

$P_8 = S_1P_7-S_2P_6+S_3P_5$

Therefore,

$P_1 = 1$

$2 = 1-2S_2 \quad \Rightarrow S_2 = -\frac {1}{2}$

$3 = 2 + \frac {1}{2}+3S_3 \quad \Rightarrow S_3 = \frac {1}{6}$

$P_4 = 3+\frac {1}{2}(2)+\frac {1}{6}(1) = \frac {25}{6}$

$P_5 = \frac {25}{6}+\frac {1}{2}(3)+\frac {1}{6}(2) = 6$

$P_6 = 6+\frac {1}{2}( \frac {25}{6})+\frac {1}{6}(3) = \frac {103}{12}$

$P_7 = \frac {103}{12} +\frac {1}{2}(6)+\frac {1}{6}(\frac {25}{6}) = \frac {221}{18}$

$P_8 = \frac {221}{18} +\frac {1}{2}(\frac {103}{12}) + \frac {1}{6}(6) = \frac {1265}{72}$

Therefore, $a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8+a_9$

$= 25 + 6 + 6 + 103 + 12 + 221 + 18 + 1265 + 72 = \boxed {1728}$

Using Newton's Identity...

But since the polynomial is of third-degree polynomial, there are 4 terms and since there are 4 terms, there are 4 numerical coefficients, so for a(n-k) = 0 for k greater than or equal to 4.

This reduces to a(n)S(7) + a(n-1)S(6) + a(n-2)S(5) + a(n-3)S(4) = 0 where from earlier, we assumed that a(n) = 1, so a(n-1) = -1, a(n-2) = -1/2, and a(n-3) = -1/6.

Where we assume that a(n) = leading coefficient of polynomial = 1 of a(3)x^3 + a(2)x^2 + a(1)x + a(0) = P(x).

After algebraic manipulations, we get that S(8) = 1265/72 and from earlier, S(7) = 221/18, S(6) = 103/12, S(5) = 6, and S(4) = 25/6. Hence, the answer.

$Using~Newton~ and~notations,~from~ the~ given~ data:-\\ P_1~=~S_1=1...............................................................................P_1=1\\ P_2~=~S_1*P_1 - 2*S_2~~~=~2,~so......S_2~=~-1/2.......................P_2=2\\ P_3~=~S_1*P_2 - S_2*P_1~+3*S_3*=~3,~so......S_3~=~1/6..................P_3=3\\ ~~~~\\ For~Greater~than~P_3,~~P_i~=~S_1*P_{i-1} - S_2*P_{i-2}+S_3*P_{i-3}.\\ So ~for~this~problem~~~\color{#3D99F6}{P_i~=~P_{i-1} +1/2*P_{i-2}+1/6*P_{i-3}.}\\ \therefore\\ P_4~=~3+1/2*2+1/6*1~=~25/6=a_1/a_2,~~~~~~~~~~~~~~~~~~~~a_1=25~~a_2=6.\\ P_5~=~25/6+1/2*3+1/6*3=~6~=~a_3,~~~~~~~~~~~~~~~~~~~~~~~~~~a_3=6.\\ P_6~=~6+1/2*25/6+1/6*3=~103/12~=~a_4/a_5,~~~~~~~~~~~~~~~~~~~a_4=103~~a_5=12.\\ P_7~=~103/12+1/2*6+1/6*25/6=~221/18~=~a_6/a_7,~~~~~~~~~~~~~~a_6=221~~a_7=18.\\ P_8~=~221/18+1/2*103/12+1/6*6=~1265/72~=~a_4-8/a_9,~~~~~~~~~~~~~~a_8=1265~~a_9=72.\\ \displaystyle \sum_{i=1}^9a_i=\Large \color{#D61F06}{1728}.$

12x12x12 is answer...