Solve the system of equations: $\left\{\begin{array}{l}x^3-y^3+(3x^2+y-2)\sqrt{y+1}-(3y^2+x-2)\sqrt{x+1}=0\\x^2+y^2+xy-7x-6y+14=0\end{array}\right.$
I used wolframalpha.com and got the solution: $(x;y)\in\left\{(2;2);\left(\dfrac{7}{3};\dfrac{7}{3}\right)\right\}$
And combining with symmetry of first equation, I guess that we can get $x=y$ from first equation.
So who can help me?
substitute for y in the second equation and solve a quadratic: $$3x^2-13x+14=0\Rightarrow(3x-7)(x-2)=0$$...etc...