2023/014) Solve in real x, y $x^3+x = y^3 + y$

We have

 $x^3+x = y^3 + y$

$=>x^3-y^3 + x - y= 0$

$=>(x-y)(x^2+xy+ y^2)  + (x - y)= 0$

$=>(x-y)(x^2+xy+ y^2+1) = 0$

$=>$ $x-y = 0$ $=>x = y$

or $x^2+xy+y^2 + 1 = 0$

but $x^2+xy+y^2  +1 = (x-\frac{y}{2})^2 + \frac{3}{4}y^2 + 1 > 1$ which does not have real solution

so solution x = y