smc

Let $p,q$, and $r$ be the roots of the polynomial. Then, $p^3-5p^2+8p-13=0$ $q^3-5q^2+8q-13=0$ $r^3-5r^2+8r-13=0$ Adding these three equations, we get $(p^3+q^3+r^3)-5(p^2+q^2+r^2)+8(p+q+r)-39=0$ $s_3-5s_2+8s_1=39$ $39$ can be written as $13s_0$, giving $s_3=5s_2-8s_1+13s_0$ We are given that $s_{k+1}=a{s}_k+b{s}_{k-1}+c{s}_{k-2}$ is satisfied for $k=2$, $3$, $....$, meaning it must be satisfied when $k=2$, giving us $s_3=a{s}_2+b{s}_1+c{s}_0$. Therefore, $a=5,b=-8$, and $c=13$ by matching coefficients. $5-8+13=\boxed{10}$.