<s>[INST] How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers: 
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
 [/INST] Subtract the two equations to get 
$(x^2 -y^2) -(x^3 -y^3)-(x-y)=0$
$\Rightarrow (x-y)(x+y-x^2-y^2 -xy-1)=0$
So, either $x-y=0$ or $x+y-x^2-y^2 -xy-1=0$


*

*For the first case, 
We have $x^3+x^2-x-1=(x-1)(x^2 -1)=(x-1)^2 (x+1)=0 \Rightarrow x=1 $ or $x=-1$
So, $x=y=1$ or $x=y=-1$

*And for the second case,
$x+y-x^2-y^2 -xy-1=-\frac{1}{2} \cdot [(x-1)^2 +(y-1)^2 +(x+y)^2]=0$ 
$\Rightarrow x=1, y=1 $ and $x=-y$, which is not possible at the same time. 
So, this case yields no solutions.
Thus the only solutions are $x=y=1$ and $x=y=-1$
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