Then $a^3=f(a)=a^3+a^3+ab+c$ yields $a^3+ab+c=0$, and $b^3=f(b)=b^3+ab^2+b^2+c$ yields $ab^2+b^2+c=0$. This implies $a^3+ab=ab^2+b^2$ and $(a+1)b^2-ab-a^3=0$. The case $a=-1$ yields $b=-1=a$ (which violates the condition of mutually distinct integers). Otherwise this quadratic equation for $b$ has the solutions $b=a$ (which violates the condition of mutually distinct integers), and $b=-a^2/(a+1) $. Now $b=-a^2/(a+1) $ is integer. For $a\ge1$, the numbers $a^2$ and $a+1$ are relative prime and $a+1\ge2$ has a prime divisor that does not divide $a$. For $a\le-3$, the numbers $a^2$ and $a+1$ are relative prime and $a+1\le-2$ has a prime divisor that does not divide $a$. There remain cases $a=0$ (contradiction to different from zero), $a=-1$ (already excluded above), and $a=-2$. Hence $a=-2$ and $b=4$, and $c=16$. One easily checks that $f(x)=x^3-2x^2+4x+16$ has all desired properties.

$ f(a)=2a^3+ab+c=a^3$ $...(A)$ $f(b)=b^3+ab^2+b^2+c=b^3$ $...(B)$ Subtract (A) from (B)and we have $ab^2+b^2-a^3-ab=0$ $<=>(b-a)(a^2+ab+b)=0. =>a^2+ab+b=0$ $<=> a^2+ab+b-1=-1$ $<=> (a+1)(a+b-1)=-1$ $=> a=-2, b=4, c=16$

Define $g(x) = f(x) - x^3$ . Therefore we have $g(x) = ax^2 + bx+ c = 0$ . Also it is easy to see that $g(a) = 0$ $ g(b) = 0$ Therefore using $Viete $ , we have after a short computation $$ a^2 + ab + b = 0 $$and $$ ab = \frac{c}{a}$$ Now we can easily get a factorization $ a^2 + ab + b -1= -1 $ Hence , we have $ (a+1) (a+b-1) = -1 $ Hence we conclude (after a short case analysis that $ (a+1) $ cannot be equal to 1 as then a will be equal to 0. $$ a = -2 , b =4 , c= 16 $$And we are done

Aryan-23 wrote: Define $g(x) = f(x) - x^3$ . Therefore we have $g(x) = ax^2 + bx+ c = 0$ . Also it is easy to see that $g(a) = 0$ $ g(b) = 0$ Therefore using $Viete $ , we have after a short computation $$ a^2 + ab + b = 0 $$and $$ ab = \frac{c}{a}$$ Now we can easily get a factorization $ a^2 + ab + b -1= -1 $ Hence , we have $ (a+1) (a+b-1) = -1 $ Hence we conclude (after a short case analysis that $ (a+1) $ cannot be equal to 1 as then a will be equal to 0. $$ a = -2 , b =4 , c= 16 $$And we are done Good