Let x=y+2. $g(y)=f(x)=y^3+5y+18,g(-y)=36-g(y).$ Therefore if f(a)+f(b)=36, then a+b=4.

Basically the same idea: the graph of the function $f(x)=x^3+mx^2+nx+p$ is symmetric across the point $A(-m/3,f(-m/3))$. In our case, $A(2,18)$. Therefore, the points $B(a,16),C(b,20)$ are also symmetric across $A$. It follows that the midpoint of the interval $(a,b)$ is $2$, that is, $a+b=4.$ P.S. I used the fact that $f$ is increasing, therefore the equations $f(x)=16,f(x)=20$ have unique real solutions.

I cant see how Rust say that " if f(a)+f(b)=36 then a+b=4 ? " wat i do is Letting $g(x)=x^3-6x^2+17x-18=(x-2)(x^2-4x+9)$ and $g(a)=-2$ , $g(b)=2$ so $g(a)=(a-2)(a^2-4a+9)=-2$ $\implies a-2=\frac{-2}{a^2-4a+9}$ similarly $b-2=\frac{2}{b^2-4b+9}$ . Add up to get $a+b-4=\frac{2(a-b)(a+b-4)}{(a^2-4a+9)(b^2-4b+9)}$ so $a+b=4$ but I dont know if $a+b\not =4$ , would there be a solution or not for $a,b$

are you sure that a,b are reals? if they can be complex,there are a pile of solutions!