12.32 Find the integer solutions of the equation $y^{3}-x^{3}=91$.
(Polish Mathematical Competition, 1949)
［Solution］The original equation can be transformed into
$$
(y-x)\left(y^{2}+x y+x^{2}\right)=13 \cdot 7 .
$$

Since for all non-zero real numbers $x$ and $y$,
$$
y^{2}+x y+x^{2}>0 \text {. }
$$

Therefore,
$$
y>x \text {. }
$$

Since 13 and 7 are prime numbers, the given equation has the following cases:
I . $\left\{\begin{array}{l}y-x=91, \\ y^{2}+x y+x^{2}=1 .\end{array}\right.$
II. $\left\{\begin{array}{l}y-x=1, \\ y^{2}+x y+x^{2}=91\end{array}\right.$.
III. $\left\{\begin{array}{l}y-x=13, \\ y^{2}+x y+x^{2}=7 .\end{array}\right.$
IV. $\left\{\begin{array}{l}y-x=7, \\ y^{2}+x y+x^{2}=13 .\end{array}\right.$

Equation sets I and III have no real solutions, and thus no integer solutions.
Solving equation sets II and IV yields four sets of solutions:
$$
\left\{\begin{array} { l } 
{ x = 5 , } \\
{ y = 6 , }
\end{array} \left\{\begin{array} { l } 
{ x = - 6 , } \\
{ y = - 5 , }
\end{array} \left\{\begin{array} { l } 
{ x = - 3 , } \\
{ y = 4 , }
\end{array} \left\{\begin{array}{l}
x=-4, \\
y=3 .
\end{array}\right.\right.\right.\right.
$$