समाधान x^12+y^2=1

$$x ^ { 12 } + y ^ { 2 } = 1$$

x को लागि हल गर्नुहोस् (complex solution)

$$x\in e^{\frac{\pi i}{6}}\sqrt[12]{1-y^{2}},\sqrt[12]{1-y^{2}},e^{\frac{\pi i}{3}}\sqrt[12]{1-y^{2}},i\sqrt[12]{1-y^{2}},e^{\frac{2i\pi }{3}}\sqrt[12]{1-y^{2}},e^{\frac{5i\pi }{6}}\sqrt[12]{1-y^{2}},-\sqrt[12]{1-y^{2}},e^{\frac{7i\pi }{6}}\sqrt[12]{1-y^{2}},e^{\frac{4i\pi }{3}}\sqrt[12]{1-y^{2}},-i\sqrt[12]{1-y^{2}},e^{\frac{5i\pi }{3}}\sqrt[12]{1-y^{2}},e^{\frac{11i\pi }{6}}\sqrt[12]{1-y^{2}}$$

y को लागि हल गर्नुहोस् (complex solution)

$$y=-\sqrt{\left(1-x^{4}\right)\left(x^{4}-x^{2}+1\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)}$$

$$y=\sqrt{\left(1-x^{4}\right)\left(x^{4}-x^{2}+1\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)}$$

x को लागि हल गर्नुहोस्

$$\left\{\begin{matrix}x=\sqrt[6]{-\sqrt{1-y^{2}}}\text{; }x=-\sqrt[6]{-\sqrt{1-y^{2}}}\text{, }&|y|=1\\x=\sqrt[12]{1-y^{2}}\text{; }x=-\sqrt[12]{1-y^{2}}\text{, }&|y|\leq 1\end{matrix}\right.$$

y को लागि हल गर्नुहोस्

$$y=\sqrt{\left(1-x^{4}\right)\left(x^{4}-x^{2}+1\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)}$$

$$y=-\sqrt{\left(1-x^{4}\right)\left(x^{4}-x^{2}+1\right)\left(\left(x^{2}+1\right)^{2}-x^{2}\right)}\text{, }|x|\leq 1$$