Polynomial

$f(x) = x^6-x^5-x^3-x^2-x$

$\quad \quad = x^2(x^4-x^3-x^2-1) + x^4-x^3 - x$

$\quad \quad = x^2(x^4-x^3-x^2-1) + (x^4-x^3 - x^2 -1) + x^2 - x + 1$

$\quad \quad = (x^2+1) P(x) + x^2 - x + 1$

Since $a$ , $b$ , $c$ and $d$ are roots of $P(x)$

$\Rightarrow P(a) = P(b) = P(c) = P(d) =0$

$f(a) = (a^2+1)P(a) + a^2 - a +1 = 0 + a^2 - a +1 = a^2 - a +1$

$f(b) = b^2 - b +1\quad f(c) = c^2 - c +1\quad f(d) = d^2 - d +1$

Therefore,

$f(a) + f(b) + f(c) + f(d) = (a^2+b^2+c^2+d^2) - (a+b+c+d) + (1+1+1+1)$

By Vieta's sums, we have:

$a+b+c+d = 1$

$a^2+b^2+c^2+d^2 = (a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd) = 1^2 - 2(-1) = 3$

$\Rightarrow f(a) + f(b) + f(c) + f(d) = 3 - 1 + 4 = \boxed {6}$

P(x) $quad =\quad { x }^{ 4 }-{ x }^{ 3 }-{ x }^{ 2 }-1\quad =\quad 0\\ \Rightarrow { x }^{ 5 }-{ x }^{ 4 }-{ x }^{ 3 }-x\quad =\quad 0\\ \Rightarrow { x }^{ 6 }-{ x }^{ 5 }-{ x }^{ 4 }-{ x }^{ 2 }\quad =\quad 0\\ so\quad f(x)\quad =\quad { x }^{ 2 }-x+1\quad for\quad all\quad x\quad \in \quad \{ a,b,c,d\} \\ \Rightarrow f(a)+f(b)+f(c)+f(d)\quad =\quad \sum { { a }^{ 2 } } -\sum { { a } } +\sum { 1 } \quad =\quad \sum { { a }^{ 2 } } -\sum { { a } } +4\\ also\quad by\quad P(x)\quad we\quad get\quad \sum { { a } } \quad =\quad 1\quad and\sum { ab } \quad =\quad -1\\ so\quad we\quad get\quad \sum { { a }^{ 2 } } \quad =\quad 3\\ \therefore f(a)+f(b)+f(c)+f(d)=3-1+4\quad =$ quad 6(ans)