Crazy system of equations

Assuming $x$ is real...

Set $f(u) = 2u^3-1$ . Then the given system is equivalent to $\begin{cases}y = f(f(x)) \\ x = f(f(y))\end{cases}$ and therefore $f(f(f(f(x)))) = x$ .

We note that $f$ is strictly increasing, so by repeated application of $f$ , we see $\begin{aligned}f(x)<x &\implies f(f(f(f(x)))) < f(f(f(x))) < f(f(x)) < f(x) < x \\ f(x)>x &\implies f(f(f(f(x)))) > f(f(f(x))) > f(f(x)) > f(x) > x\end{aligned}$ so that we must have $f(x) = x$ .

That is, $2x^3 - 1 = x \\ 2x^3 - x - 1 = 0 \\ (x-1)(2x^2+2x+1) = 0$ showing that the only real solution is $x=1$ .

This gives $y = f(f(1)) = 1$ and therefore $\boxed{x+y=2}$

We can clearly see that f (x)=f (inverse) of y {where f (x)=2x^3-1}. Similarly f (y)=f (inverse) of x.
Thus f (f (f (x)))=x. Now (f (x))<x if x <1. And f (x)>x if x>1 thus.
f (f....f (x)))..))<x if x <1 and so is greater . than x part is proved.
And thus only soln for x is x=1.
And for this y=1 satisfies.
thus x+y=2

X=Y=1 is a solution.