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Subtracting the given equations, we obtain $y-x+x^2-y^2=x^3-y^3$, which is equivalent to $$(x-y)(-1+x+y)=(x-y)(x^2+xy+y^2),$$ implying that either $x-y=0$ or $x+y-1=x^2+xy+y^2$.

In the latter case, $x+y-1=x^2+xy+y^2$ is equivalent to $$y^2+(x-1)y+x^2-x+1=0.(\ast )$$ This is a quadratic equation in $y$. Its discriminant is $$D=(x-1{)}^2-4(x^2-x+1)=-3x^2+2x-3,$$ which is a quadratic polynomial in $x$ – its discriminant in turn is $-32<0$, so the starting discriminant is $D<0$ for every real number $x$, which implies that the equation $(\ast )$ does not have any real solution $y$. Hence there are no solutions in this case.

In the remaining case, $y=x$ so both equations of a given system are reduced to $x+x^2=x^3$, i.e. $x(x^2-x-1)=0$. This equation has three solutions $x_1=0$, $x_{2,3}=\frac{1\pm \sqrt{5}}{2}$.

The only solutions are $(x,y)\in \{(0,0),(\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}),(\frac{1-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})\}$.