Tìm x biết: x^3 – y^3 = 91.

x³ – y³ = 91

⇔ (x – y)(x² + xy+ y²) = 91

Do x – y < x² + xy+ y² và x² + xy+ y² > 0 nên ta có 2 trường hợp:

TH1: \(\left\{ \begin{array}{l}x - y = 1\\{x^2} + xy + {y^2} = 91\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}x = y + 1\\{\left( {y + 1} \right)^2} + y\left( {y + 1} \right) + {y^2} = 91\end{array} \right.\)

⇒ 1 + 2y + y² + y + y² + y² = 91

⇔ 3y² + 3y – 90 = 0

⇔ \(\left[ \begin{array}{l}y = 5\\y = - 6\end{array} \right.\)

Suy ra: \(\left[ \begin{array}{l}x = 6\\x = - 5\end{array} \right.\)

TH2: \(\left\{ \begin{array}{l}x - y = 7\\{x^2} + xy + {y^2} = 13\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}x = y + 7\\{\left( {y + 7} \right)^2} + y\left( {y + 7} \right) + {y^2} = 13\end{array} \right.\)

⇔ 3y² + 21y + 36 = 0

⇔ \(\left[ \begin{array}{l}y = 3\\y = - 4\end{array} \right.\)

Suy ra: \(\left[ \begin{array}{l}x = 4\\x = 3\end{array} \right.\)

Vậy (x; y) ∈ {(6; 5); (-5; -6); (4; -3); (3; -4)}.