$f(x)f(-x)=(-x^3+ax^2-bx+c)(x^3+ax^2+bx+c)$ $=-x^6+(a^2-2b)x^4+(2ac-b^2)x^2+c^2$, so $g(x)=x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2$. And $2b-a^2=b$, $b^2-2ac=c$, $-c^2=a$. We can solve these: $b=a^2$, $a=-c^2$, so $(a, b, c)=(-c^2, c^4, c)$. $f(1)$ gives $-c^2+c^4+c+1=0$ which factorizes as $(c+1)(c^3-c^2+1)$. So $c=a=-1, b=1$. So then $a^{2013}+b^{2013}+c^{2013}=-1+1-1=-1$.

Can you please say why did you decided to compute $f(x)f(-x)$?

tenniskidperson3 wrote: $f(x)f(-x)=(-x^3+ax^2-bx+c)(x^3+ax^2+bx+c)$ $=-x^6+(a^2-2b)x^4+(2ac-b^2)x^2+c^2$, so $g(x)=x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2$. And $2b-a^2=b$, $b^2-2ac=c$, $-c^2=a$. We can solve these: $b=a^2$, $a=-c^2$, so $(a, b, c)=(-c^2, c^4, c)$. $f(1)$ gives $-c^2+c^4+c+1=0$ which factorizes as $(c+1)(c^3-c^2+1)$. So $c=a=-1, b=1$. So then $a^{2013}+b^{2013}+c^{2013}=-1+1-1=-1$. No need to Calculate $f(x)f(-x)$...Just use Viete's Relations to get $-c^2+c^4+c+1=0$ .. Then it's straight forward

Btw. Here's my solution:-- Note that $g(1) = f(1) = 0$, so $1$ is a root of both $f(x)$ and $g(x)$. Let $p$ and $q$ be the other two roots of f(x), so $p^2$ and $q^2$ are the other two roots of $g(x)$. We then get $ pq =-c$ and$p^2.q^2 = -a$, so $a = -c^2$. Also, $(-a)^2$ = $(p+q +1)^2 = $$p^2 +q^2+1 + 2(pq +p+q) = -b+ 2b = b$.Therefore $b = c^4$. Since $f(1) = 0$we therefore get $1 + c - c^2 + c^4 = 0$ . Factorising, we get $(c + 1)(c^3 - c^2 + 1)= 0$. Note that $c^3 - c^2 + 1 = 0$ has no integer root and hence $c = -1, b = 1, a = -1$. Therefore $a^{2013} + b^{2013} + c^{2013} = -1$ Q.E.D

Disclaimer: I haven’t given the explanation behind my solution but nonetheless if anyone want to look they may..

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By Vieta's Formulas on the first polynomial, we have that $\sigma_1=-a$, $\sigma_2=b$, and that $\sigma_3=-c$. Using it again on the second polynomial, we have that \begin{align*} \rho_2=(-a)^2-2b&=-b \\ a^2&=b \end{align*}\begin{align*} \sigma_3^2=(-c)^2=-a \end{align*}\begin{align*} r^2s^2+s^2t^2+t^2r^2=b^2-2(-a)(-c)&=c \\ a^4&=(2a+1)c \\ c&=\frac{a^4}{2a+1} \\ c&=\frac{c^8}{-2c^2+1} \\ c^7&=1-2c^2 \\ \end{align*}The only possible value is $c=-1$, meaning that $a=-1$ and $b=1$. So the answer is $\boxed{-1}$. We also don't need the fact that $a$, $b$, and $c$ are integers.

Assume the roots of the Polynomial $f(x)=x^3+ax^2+bx+c$ are $\alpha,\beta,\gamma.$ and the roots of the polynomial $g(x)=x^3+bx^2+cx+a$ are $\alpha^2,\beta^2,\gamma^2.$ Given $f(1)=0$ and the roots of $g(x)=0$ are the squares of roots of $f(x)=0.$ we have to find the value of $a^{2013}+b^{2013}+c^{2013}.$ By Vieta's Relation we get, $$\alpha+\beta+\gamma=-a\implies\boxed{\sum_{cyc}{}\alpha^2=a^2-2b}$$$$\alpha\beta+\beta\gamma+\gamma\alpha=b\implies\boxed{\sum_{cyc}{}\alpha^2\beta^2=b^2-2ac}$$$$\alpha\beta\gamma=-c\implies\boxed{\alpha^2\beta^2\gamma^2=c^2}$$Also we get, $\boxed{2a+c=-1}$ from $f(1)=0.$ Therefore we get, $$g(x)=x^3+(2b-a)x^2+(b^2-2ac)x-c^2=x^3+bx^2+cx+a$$comparing these we get, $$a=b, a^2-2ac=c,c^2=-a$$From $2a+c=-1\implies 4a^2+4a+1=\underbrace{(4a+1)}_{not possible }(a+1)\implies\boxed{a=b=-1}$ and $\boxed{c=1}$ Therefore we get, $$a^{2013}+b^{2013}+c^{2013}=\boxed{-1}$$Vietas' Relation!!!!

f(x) = x³ + ax² + bx + c and g(x) = x³ + bx² + cx + a where a,b and c are integers with c ≠ 0 f(1) = 0 i.e., a + b + c = -1 Let p,q,r be the 3 roots of the polynomial f(x) and p²,q² and r² that of polynomial g(x). From the first polynomial f(x), we've p + q + r = -a (Sum of the roots) ———(1) pq + qr + rp = b (Sum of the roots taken 2 at a time) ———(2) pqr = -c (Product of the roots) ———(3) From the second polynomial g(x), we've p² + q² + r² = -b (Sum of the roots) (p + q + r)² - 2(pq + qr + rp) = -b On plugging the values of (p + q + r) and (pq + qr + rp) from equation (1) and (2), we get a² = b p²q²r² = -a (Product of the roots) a = -c² But, a + b + c = - 1 c⁴ - c² + c + 1 = 0 (since b = a² = c⁴) (c + 1)(c³ - c² + 1) = 0 Therefore, c = -1 since c³ - c² + 1 = 0 has no integral roots. a = -1, b = 1 and c = - 1 So, a^2013 + b^2013 + c^2013 = -1 + 1 - 1 = -1 QED