Bashing unavailable - part 2

Using Newton's Identity where:

$\begin{cases}\\ a_nS_1 + an_1 = 0\\ a_nS_2 + an_1 S_1 + 2an_2 = 0\\ a_nS_3+ an_1 S_2 + an_2 S_1 + 3an_3 = 0\\ a_nS_4 + an_1 S_3 + an_2 S_2 + an_3 S_1 + 4an_4 = 0\\ a_nS_5 + an_1 S_4 + an_2 S_3 + an_3 S_4 + an_4 S_5 + 5an_5 = 0 \end{cases}$

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

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 manipulation, we get that $S(5) = 6$ and from earlier, $S(4) = \frac{25}{6}$ . Hence, the answer: $\large{6+6+25=\boxed{37}}$

$Using~Newton~ and~notations,~from~ the~ given~ data:-\\ P_1~=~S_1=1...............................................................................P_1=1\\ P_2~=~S_1*P_1 - 2*S_2~~~=~2,~so......S_2~=~-1/2.......................P_2=2\\ P_3~=~S_1*P_2 - S_2*P_1~+3*S_3*=~3,~so......S_3~=~1/6..................P_3=3\\ ~~~~\\ For~Greater~than~P_3,~~P_i~=~S_1*P_{i-1} - S_2*P_{i-2}+S_3*{i-3}.\\ So ~for~this~problem~~~\color{#3D99F6}{P_i~=~P_{i-1} +1/2*P_{i-2}+1/6*P_{i-3}.}\\ \therefore\\ P_4~=~3+1/2*2+1/6*1~=~25/6=a_1/a_2,.................................a_1=25~~a_2=6.\\ P_5~=~25/6+1/2*3+1/6*3=6~=~a_3\\ \therefore~~a_1+a_2+a_3~=~\Large \color{#D61F06}{37}.\\$ .