Polynomial problem with unknown coefficients $a, b, c$

Question

$p(x)=x^3 +ax^2+bx+c$, where $a,b,c$ are distinct non-zero integers. Suppose $p(a)=a^3$ and $p(b)=b^3$, find $p(13)$.

Ok so I've gotten $a^3-b^3=a^3-b^3+a^3-ab^2+ab-b^2$, So $(a-b)[a(a+b)+b]=0$, so $b=-\frac{a^2}{a+1}$ and then getting $c=-\frac{a^4}{a+1}$ does not help much I guess. Will someone tell me how I should be thinking about this question? Thanks.

Answer 1

Put $n=a+1$, then: $$b={-a^2\over a+1} = -{(n-1)^2\over n} = -{n^2-2n+1\over n} = -n+2-{1\over n} \in\mathbb{Z}$$

So $$n\mid 1\implies n =\pm 1 \implies a=0 \vee a= -2 $$ Now if $a=0$ then $b=0$ and this can't be, so $a=-2$ and $b=4$ and $c= 16$.

Answer 2

Since $b=-\frac{a^2}{a+1}$, we see that $$c=-a^3-ab=-a^3+\frac{a^3}{a+1}=-\frac{a^4}{a+1}$$ and since $gcd(a^4,a+1)=1,$ we obtain:

$a+1\in\{-1,1\}$ and since $a\neq0$, we obtain $a=-2$, $c=16$, $b=4$, $$f(x)=x^3-2x^2+4x+16$$ and $$f(13)=13^3-2\cdot13^2+4\cdot13+16=1927.$$