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Algebra Difficulty 5.2 AIME, harder Find the answer

3. Let the function f(x)=x3+ax2+bx+c(x∈R)f(x)=x^{3}+a x^{2}+b x+c(x \in \mathbf{R}), where a,b,ca, b, c are distinct non-zero integers, and f(a)=a3,f(b)=f(a)=a^{3}, f(b)= b3b^{3}, then a+b+c=a+b+c= \qquad .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Given that a,ba, b are the roots of the equation ax2+bx+c=0a x^{2}+b x+c=0 ⇒{a+b=−ba,ab=ca⇒b(1+1a)=−a\Rightarrow\left\{\begin{array}{l}a+b=-\frac{b}{a}, \\ a b=\frac{c}{a}\end{array} \Rightarrow b\left(1+\frac{1}{a}\right)=-a\right.
⇒b=−a2a+1=1−a−1a+1∈Z⇒a=−2,b=4,c=16.  \Rightarrow b=\frac{-a^{2}}{a+1}=1-a-\frac{1}{a+1} \in \mathbf{Z} \Rightarrow a=-2, b=4, c=16 \text {. }

Therefore, a+b+c=18a+b+c=18.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.