So $c = -(a+b+1)$ . Then $f(x) = (x-1)(x^2 + (a+1)x + (a+b+1))$ and $g(x) = (x-1)(x^2 + (b+1)x - a)$ Let $1,u,v$ be the roots of $f$ . It follows $-(b+1) = u^2+v^2 = (u+v)^2 - 2uv = (a+1)^2 - 2(a+b+1)$ , whence $b = a^2$ , and $-a = u^2v^2 = (a+b+1)^2$ , whence $a = -c^2$ . It follows $0 = 1+a+b+c = 1-c^2 +c^4 + c = (c+1)(c^3-c^2+1)$ , whence $c=-1$ , and so $a=-1$ and $b=1$ . Therefore $\boxed{a^{2013} + b^{2013} + c^{2013} = -1}$

by relationship between a,b and c we get

a+b+c=-1

b=a^2

-a=c^2

therefore c^4-c^2+c+1=0

its 1 root is -1

therefore a=c=-1 and b=1

hence answer=-1+1-1=-1

considering the roots of f(x)=0, as p,q and r- we see the roots of g(x)=0 as 1/p, 1/q and 1/r. Also, i) p+q+r=-1 ii) 1/p, 1/q and 1/r is same as p^2, q^2 and r^2. iii) fact that pqr is not 0. two of them are -1 and remaining is 1.