Giải thích các bước giải:
$+)x=0\to 3^0-y^3=1\to y=0$
$+) x>0\to y^3+1=3^x\to (y+1)(y^2-y+1)=3^x$
$\to\begin{cases}y+1=3^m\\ y^2-y+1=3^n\end{cases}$
Gọi $(y+1,y^2-y+1)=d$
$\to (y+1)^2-(y^2-y+1)\quad\vdots\quad d$
$\to 3y\quad\vdots\quad d$
$\to 3(y+1)-3y\quad\vdots\quad d$
$\to 3\quad\vdots\quad d$
$\to d=3$ hoặc $d=1$
$+)d=3\to (3^m,3^n)=3\to m=1$ hoặc $n=1$
Nếu $m=1\to y=2\to x=9\to x=2$
Nếu $n=1\to y^2-y+1=3\to y\in\{-1,2\}\to x\in\{0,9\}\to y=2,x=2$
$+)d=1\to (3^m,3^n)=1\to m=0$ hoặc $n=0$
Nếu $m=0\to y+1=1\to y=0\to 3^x=1\to x=0$
Nếu $n=0\to y^2-y+1=1\to y\in\{0,1\}\to 3^x\in\{1,2\}\to y=0,x=0$
