Đáp án:
$(x,y)=(\frac{3}{2},\frac{1}{2}); (0,-1)$
Giải thích các bước giải:
$x^{3}-y^{3}-6x^{2}+3y^{2}-6y+15x-10=0\\
\rightarrow (x^{3}-6x^{2}+12x-8)+3(x-2)=(y^{3}-3y^{2}+3y-1 )+3(y-1)\\
\rightarrow (x-2)^{3}+3(x-2)=(y-1)^{3}+3(y-1)\\
\rightarrow ((x-2)^{3}-(y-1)^{3})+(3(x-2)-3(y-1))=0\\
\rightarrow ((x-2)-(y-1))((x-2)^{2}+(x-2)(y-1)+(y-1)^{2})+3((x-2)-(y-1))=0\\
\rightarrow (x-y-1)((x-2)^{2}+(x-2)(y-1)+(y-1)^{2}+3)=0\\
\rightarrow x-y-1=0\\
\rightarrow x=y+1\\
\rightarrow y^{2}+(y+1)^{2}-2(y+1)+y-2=0\\
\rightarrow 2y^{2}+y-1=0\\
\rightarrow (2y-1)(y+1)=0\\
\rightarrow y=\frac{1}{2} \rightarrow x=y+1=\frac{3}{2}\\
\quad \quad \text{Hoặc } y=-1 \rightarrow x=y+1=0$
