Giải thích các bước giải:
Từ hệ ta suy ra:
$x^3-2y^3=(x+4y)(6x^2-19xy+15y^2)$
$\to x^3-2y^3=6x^3+5x^2y-61xy^2+60y^3$
$\to 5x^3+5x^2y-61xy^2+62y^3=0$
$\to \left(x-2y\right)\left(5x^2+15xy-31y^2\right)=0$
$\to x-2y=0$
$\to x=2y$
$\to 6\cdot (2y)^2-19\cdot 2y\cdot y+15y^2=1$
$\to y^2-1=0$
$\to y=\pm1\to x=\pm2$
Hoặc $5x^2+15xy-31y^2=0$
$\to x=\dfrac{-15y\pm13\sqrt{5}y}{10}$
Nếu $x=\dfrac{-15y+13\sqrt{5}y}{10}$
$\to 6\cdot (\dfrac{-15y+13\sqrt{5}y}{10})^2-19\cdot \dfrac{-15y+13\sqrt{5}y}{10}\cdot y+15y^2=1$
$\to \left(\dfrac{1077}{10}-\dfrac{481}{2\sqrt{5}}\right)y^2-1=0$
$\to y=\pm\dfrac{\sqrt{2}\cdot \:5^{\dfrac{3}{4}}\left(1077\sqrt{5}+2405\right)\sqrt{1077\sqrt{5}-2405}}{15620}\to x$
Nếu $x=\dfrac{-15y-13\sqrt{5}y}{10}$
$\to 6\cdot (\dfrac{-15y-13\sqrt{5}y}{10})^2-19\cdot \dfrac{-15y-13\sqrt{5}y}{10}\cdot y+15y^2=1$
$\to \dfrac{\left(1077+481\sqrt{5}\right)y^2}{10}=1$
$\to y=\pm\sqrt{\dfrac{5\left(1077-481\sqrt{5}\right)}{1562}}\to x$
