Løs x^4+y^4=16

$${ x }^{ 4 } + { y }^{ 4 } =16$$

Løs for x (complex solution)

$$x=i\sqrt[4]{16-y^{4}}$$

$$x=\sqrt[4]{16-y^{4}}$$

$$x=-\sqrt[4]{16-y^{4}}$$

$$x=-i\sqrt[4]{16-y^{4}}$$

Løs for y (complex solution)

$$y=i\sqrt[4]{16-x^{4}}$$

$$y=\sqrt[4]{16-x^{4}}$$

$$y=-\sqrt[4]{16-x^{4}}$$

$$y=-i\sqrt[4]{16-x^{4}}$$

Løs for x

$$\left\{\begin{matrix}x=\sqrt{-\sqrt{4-y^{2}}}\sqrt[4]{y^{2}+4}\text{; }x=-\sqrt{-\sqrt{4-y^{2}}}\sqrt[4]{y^{2}+4}\text{, }&|y|=2\\x=\sqrt[4]{16-y^{4}}\text{; }x=-\sqrt[4]{16-y^{4}}\text{, }&|y|\leq 2\end{matrix}\right,$$

Løs for y

$$\left\{\begin{matrix}y=\sqrt{-\sqrt{4-x^{2}}}\sqrt[4]{x^{2}+4}\text{; }y=-\sqrt{-\sqrt{4-x^{2}}}\sqrt[4]{x^{2}+4}\text{, }&|x|=2\\y=\sqrt[4]{16-x^{4}}\text{; }y=-\sqrt[4]{16-x^{4}}\text{, }&|x|\leq 2\end{matrix}\right,$$