JOMO 6, Short 9

If we reflect the curve $y=x^4+2x^3-21x^2-21x-40$ about the line y=x the point $(x,y)=(t,t^4+2x^3-21x^2-21x-40)$ is reflected to the point $(y,x)=(t^4+2x^3-21x^2-21x-40,t)$ . So the equation for the curve which we get from the reflection is $x=y^4+2y^3-21y^2-21y-40$ , which is the other given curve. That means that these two curves only have points in common which lay on the line y=x and therefore all solutions are of the form (c,c). So if we sum up the x coordinates and the y coordinates and take the difference, we get zero!

I don't know if this method is correct or not but here it goes.

Let $f(x)=x^4+2x^3-21x^2-21x-40$ .

Now, we can see that, in the given problem, the two curves are $y=f(x)$ and $x=f(y)$ . Since we are dealing with the points of intersection of the curves, both equations of curves satisfies these points. Now,

$y=f(x)\implies f(y)=f(f(x))=(f\circ f)(x) \implies \boxed{x=(f\circ f)(x)}$

Similarly, we can get that $\boxed{y=(f\circ f)(y)}$

From these two observations, we can clearly see that the points that are the intersection of the curves will be of the form $(c,c)$ where $c\in \text{Dom}(f\circ f)$ and the domain is limited to the coordinates of the intersection points since we got those two results only when we considered the intersection points. Now, for each intersection point the difference between $x$ and $y$ coordinate is $0$ and thus for all the points, the required result will be $\boxed{0}$ .

Here, $\circ$ denotes the composition of functions, i.e., $(f\circ g)(x)=f(g(x))$ , where $f,g$ are 2 functions.