Earthquake in my mind!

first lets compute $p_{11}$ . note that $s_1=s_2=s_3=0$ , hence giving us $p_1=p_2=p_3=0$ where $s_n$ is the nth symmetric sum and $p_n$ the nth power sum. we get that $\begin{array}{c}.x^5=-7x+1\\ x^{10}=49x^2-14x+1\\ x^{11}=49x^3-14x^2+x\\ \sum(x^{11})=49\sum(x^3)-14\sum(x^2)+\sum(x)=49p_3-14p_2+p_1=\boxed{0}\end{array}$ since one part of the product is 0, the whole thing becomes $\boxed{0}$

Here,

$e_1=\sum{a}=0$

$e_2=\sum{ab}=0$

$e_3=\sum{abc}=0$

$e_4=\sum{abcd}=0$

$e_5=\sum{abcde}=abcde=1$

Let's calculate $P_8$ first.

$P_8=e_1P_7-e_2P_6+e_3P_5-e_4P_4+e_5P_3$

Since, $e_1,e_2,e_3,e_4$ are all $0$ and $e_5$ being $1$ , we get $P_8=P_3$

$P_3=e_1P_2-e_2P_1+e_3P_0$

Since, $e_1,e_2,e_3$ are all $0$ , $P_3=0$

$\implies$ $P_8=0$

Notice, $P_8$ is in multiplication with $P_{11}$ .

Hence our answer is $\boxed{0}$ .