Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} \right) + f\left( 0 \right) + f\left( {\frac{1}{2}} \right) = \_\_\_\_\_$
My approach is as follow, as it is an implicit function we need to find the roots is $f(x)$.
We end up getting $(f(x)-1)(f(x)+x)(f(x)-x)=0$, so we end up getting three function viz.
$f(x)=1$; $f(x)=-x$ & $f(x)=x$ but how do we proceed further

$y=f(x). y^3-y^2-x^2y+x^2=0$. Max $y$ is $1$.  Min $y$ is $0$. Find $f(-4/3)+f(0)+f(1/2)$
$g(x,y)= (y-x)(y+x)(y-1)=0$.
The constraint requires only certain $y$ values for a given $x$.
$x=0\implies y\in\{0,1\}$
$x=1/2\implies y^3-y^2-y/4+1/4=0.\implies (2y-1)(2y+1)(y-1)=0\implies y\in\{1/2,-1/2,1\}$
$x=-4/3\implies y^3-y^2-16y/9+16/9=0\implies (9y^2-16)(y-1)=(3y-4)(3y+4)(y-1)=0$
$\implies y \in \{4/3, -4/3,1\}$
I'm getting only 18 possibilities, but those should be pared down. Not sure how to use the min and max values. $f(x)\equiv1$ doesn't allow a minimum value of $0$, so the $1$'s can be ignored.
Consistent definition of $f(x)$ requires consistently using $f(x)\equiv x$ or $f(x)\equiv -x$. So, combinations are reduced still further to $\{-5/6, 5/6\}$
