75th Romanian Mathematical Olympiad

a. If $(x,y)\in A$, then $y=-x-1$. We get $$x^3+y^3+1=x^3+(-x-1{)}^3+1=x^3-(x+1{)}^3+1=x^3-(x^3+3x^2+3x+1)+1=-3x^2-3x=3x(-x-1)=3xy,$$ so $(x,y)\in B$.

b. If $(x,y)\in B$, then $x^3+y^3-3xy+1=0$, so $$(x+y{)}^3-3xy(x+y)-3xy+1=0,$$ or $$(x+y+1)((x+y{)}^2-(x+y)+1)-3xy(x+y+1)=0,$$ or $$(x+y+1)(x^2-xy+y^2-x-y+1)=0.$$ In order to have $(x,y)\in B\setminus A$ it is necessary that $x^2-xy+y^2-x-y+1=0$, whence $$(x-y{)}^2+(x-1{)}^2+(y-1{)}^2=0,$$ which means $x=y=1$. Since $(1,1)\notin A$, it follows that $B\setminus A=\{(1,1)\}$.

Alternative solution.

It is well known the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$ We get $$x^3+y^3+1-3xy=(x+y+1)(x^2+y^2+1-xy-x-y).$$ Thus, if $(x,y)\in A$, then $(x,y)\in B$. Then, if $(x,y)\in B\setminus A$, it follows that $x^2+y^2+1-xy-x-y=0$ and, after that, we finish the proof as for the first solution.

Alternative solution for b).

We notice that $(1,1)\in B\setminus A$. Consider $(x,y)\in B$, so $x=1+a,y=1+b$. Then $$a^3+b^3+3a^2+3b^2-3ab=0,$$ or $$(a+b)(a^2-ab+b^2)+3(a^2-ab+b^2)=0,$$ so $$(a+b+3)(a^2-ab+b^2)=0.$$ If $a+b+3=0$, then $x+y+1=0$ and $(x,y)\in A$. If $a^2-ab+b^2=0$, then $a=b=0$ and $(x,y)\in B\setminus A$. Thus the set $B\setminus A$ contains only the element $(1,1)$.