For which I came here

We have:

$f(x)=x^{3}+ax^{2}+bx+c$ ; $g(x)=x^{3}+bx^{2}+cx+a$ where $a,b,c\in\mathbb{I}$ and $c\neq0$

Also, since $0=f(1)=1+a+b+c=g(1)$ we have $1$ as the common root of both the polynomials.

Now, Let's write down the expressions we get by Vieta's Formulae considering $\alpha$ and $\beta$ as the other two roots for $f(x)$ and $\alpha^{2}$ and $\beta^{2}$ as the other roots for $g(x)$

$1+\alpha+\beta=-a$ ........... $(i)$

$\alpha\beta+\alpha+\beta=b$ .................. $(ii)$

$\alpha\beta=-c$ ................. $(iii)$

$1+\alpha^{2}+\beta^{2}=-b$ ............... $(iv)$

$\alpha^{2}\beta^{2}=-a$ ........... $(v)$

These are all we need.

Notice, from $(iii)$ and $(v)$ : $c^{2}=-a$

also, $(iv)$ can be written as: $(1+\alpha+\beta)^{2}-2(\alpha+\beta+\alpha\beta)=-b$ ............... $(vi)$

and $2(\alpha+\beta+\alpha\beta)=2b$ ..........from $(ii)$

Thus, by putting respective values in $(vi)$ we get:

$a^{2}=b$

So: $a^{2}=b=-c$

We also have $1+a+b+c=0$ .

So: $a=-1$

Our required expression: $\large{a^{2013}+b^{2013}+c^{2013}=a^{2013}=\boxed{-1}}$

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