แก้โจทย์ 25x^2/y-16y^2/x=xy

$$\frac { 25 x ^ { 2 } } { y } - \frac { 16 y ^ { 2 } } { x } = x y$$

หาค่า x

$$x=-\frac{\sqrt{3}i\left(y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}\right)^{-\frac{1}{3}}y^{4}}{150}+\frac{\sqrt{3}i\sqrt[3]{y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}}}{150}-\frac{\sqrt[3]{y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}}}{150}+\frac{y^{2}}{75}-\frac{\left(y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}\right)^{-\frac{1}{3}}y^{4}}{150}$$

$$x=\frac{\sqrt[3]{y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}}+y^{2}+\left(y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}\right)^{-\frac{1}{3}}y^{4}}{75}$$

$$x=\frac{\sqrt{3}i\left(y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}\right)^{-\frac{1}{3}}y^{4}}{150}-\frac{\sqrt{3}i\sqrt[3]{y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}}}{150}-\frac{\sqrt[3]{y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}}}{150}+\frac{y^{2}}{75}-\frac{\left(y^{6}+300\sqrt{3y^{9}+202500y^{6}}+135000y^{3}\right)^{-\frac{1}{3}}y^{4}}{150}\text{, }y\neq 0$$

หาค่า y

$$y=-\frac{\sqrt{3}i\left(-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}\right)^{-\frac{1}{3}}x^{4}}{96}+\frac{\sqrt{3}i\sqrt[3]{-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}}}{96}-\frac{\sqrt[3]{-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}}}{96}-\frac{x^{2}}{48}-\frac{\left(-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}\right)^{-\frac{1}{3}}x^{4}}{96}$$

$$y=\frac{\sqrt[3]{-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}}-x^{2}+\left(-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}\right)^{-\frac{1}{3}}x^{4}}{48}$$

$$y=\frac{\sqrt{3}i\left(-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}\right)^{-\frac{1}{3}}x^{4}}{96}-\frac{\sqrt{3}i\sqrt[3]{-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}}}{96}-\frac{\sqrt[3]{-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}}}{96}-\frac{x^{2}}{48}-\frac{\left(-x^{6}+240\sqrt{129600x^{6}-3x^{9}}+86400x^{3}\right)^{-\frac{1}{3}}x^{4}}{96}\text{, }x\neq 0$$