Let $f\left(x\right)=x^6-2x^5+x^3+x^2-x-1$ and $g\left(x\right)=x^4-x^3-x^2-1$ be two polynomials. Let $a,b,c$ and $d$ be the roots of $g\left(x\right)=0$. Then, the value of 

$f\left(a\right)+f\left(b\right)+f\left(c\right)+f\left(d\right)$ is

Given,

$f\left(x\right)=x^6-2x^5+x^3+x^2-x-1$

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

$\Rightarrow f\left(x\right)=\left(x^2-x\right)g\left(x\right)+2x^2-2x-1$

$\therefore f\left(a\right)=2a^2-2a-1$

$\therefore \mathrm{\Sigma }f\left(a\right)=f\left(a\right)+f\left(b\right)+\left(c\right)+f\left(d\right)$

$=2\mathrm{\Sigma }{a}^2-2\mathrm{\Sigma }a-\mathrm{\Sigma }1$

$=2\left({\left(\mathrm{\Sigma }a\right)}^2-2\mathrm{\Sigma }ab\right)-2\mathrm{\Sigma }a-4$

$\because a,b,c,d$ are roots of equation 

$g\left(x\right)=x^4-x^3-x^2-1=0$

then $\mathrm{\Sigma }a=1, \mathrm{\Sigma }ab=-1$

$\therefore \mathrm{\Sigma }f\left(a\right)=2\left[{\left(1\right)}^2-2\left(-1\right)\right]-2\left(1\right)-4$

$=6-2-4=0$