Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$
When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$.
So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$.
Now this also follows for all roots of $Q(x)$
Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]
Now by [Vieta's](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas) we know that $-z_4-z_3-z_2-z_1=-1$,
so by [Newton's Sums](https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums) we can find $z_1^2+z_2^2+z_3^2+z_4^2$
$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$
$(1)(s_2)+(-1)(1)+2(-1)=0$
$s_2-1-2=0$
$s_2=3$
So finally
$P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$