z=x^3y+xy^3.x=scost.y=se^t шешу

$$z = x ^ { 3 } y + x y ^ { 3 } . x = s \cos t . y = s e ^ { t }$$

x мәнін табыңыз

$$\left\{\begin{matrix}\\x=\frac{18^{\frac{2}{3}}\left(\sqrt[3]{y\sqrt{3\left(4y^{8}+27z^{2}\right)}+9z|y|}+\sqrt[3]{-y\sqrt{3\left(4y^{8}+27z^{2}\right)}+9z|y|}\right)}{18\sqrt[3]{y|y|}}\text{, }&\text{unconditionally}\\x=-\frac{2^{\frac{2}{3}}\sqrt[3]{\frac{y\sqrt{3\left(4y^{8}+27z^{2}\right)}+9z|y|}{9y|y|}}}{2}\text{, }&\frac{y^{6}}{27}+\frac{z^{2}}{4y^{2}}=0\text{ and }y\neq 0\end{matrix}\right.$$

y мәнін табыңыз

$$\left\{\begin{matrix}\\y=\frac{18^{\frac{2}{3}}\left(\sqrt[3]{x\sqrt{3\left(4x^{8}+27z^{2}\right)}+9z|x|}+\sqrt[3]{-x\sqrt{3\left(4x^{8}+27z^{2}\right)}+9z|x|}\right)}{18\sqrt[3]{x|x|}}\text{, }&\text{unconditionally}\\y=-\frac{2^{\frac{2}{3}}\sqrt[3]{\frac{x\sqrt{3\left(4x^{8}+27z^{2}\right)}+9z|x|}{9x|x|}}}{2}\text{, }&\frac{x^{6}}{27}+\frac{z^{2}}{4x^{2}}=0\text{ and }x\neq 0\end{matrix}\right.$$