Solve-the-system-of-equations-x-4-2x-y-y-4-x-2-y-2-3-3-

Question Number 162181 by bobhans last updated on 27/Dec/21

Solve the system of equations

\begin{matrix} x^{4} - 2 x + y = y^{4} \\ {(x^{2} - y^{2})}^{3} = 3 \end{matrix}}

Commented by benhamimed last updated on 27/Dec/21

(x^{2} + y^{2}) (x^{2} - y^{2}) - 2 x + y = 0

\sqrt[3]{3} (x^{2} + y^{2}) - 2 x + y = 0

x^{2} + y^{2} - \frac{2}{\sqrt[3]{3}} x + \frac{1}{\sqrt[3]{3}} y = 0

{(x - \frac{1}{\sqrt[3]{3}})}^{2} + {(y + \frac{1}{2 \sqrt[3]{3}})}^{2} = \frac{1}{\sqrt[3]{6}} + \frac{1}{4 \sqrt[3]{6}}

{(x - \frac{\sqrt[3]{6}}{3})}^{2} + {(y + \frac{\sqrt[3]{6}}{12})}^{2} = \frac{5 \sqrt[3]{36}}{24}

l e s s o l u t i o n d e s y s t e m s o n t l^{'} e n s e m b l e

d e s p o i n t s q u i f o r m e n t u n c e r c l e (C)

d e c e n t r e (\frac{\sqrt[3]{6}}{3}; \frac{- \sqrt[3]{6}}{12}) e t d e m i r a y o n \frac{\sqrt{30} . \sqrt[3]{6}}{12}

Answered by cortano last updated on 28/Dec/21

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