how-to-solving-x-3-y-3-4-and-x-y-1-

Question Number 76246 by benjo last updated on 25/Dec/19

how to solving x^{3} + y^{3} = 4 and

x \times y = 1 ?

Answered by john santuy last updated on 25/Dec/19

l e t u = x^{3} a n d v = y^{3} . t h e n w e h a v e u + v = 4

a n d u^{3} \times v^{3} = 1 . s o i t h i n k y o u r c a n s o l v i n g

t h i s p r o b l e m .

Commented by benjo last updated on 25/Dec/19

thanks you sir

Answered by MJS last updated on 25/Dec/19

y = \frac{1}{x}

x^{3} + \frac{1}{x^{3}} = 4

x^{6} - 4 x^{3} + 1 = 0

let x = \sqrt[3]{t}

t^{2} - 4 t + 1 = 0

t = 2 \pm \sqrt{3}

\Rightarrow

x_{1} = \sqrt[3]{2 - \sqrt{3}}; x_{2} = \sqrt[3]{2 + \sqrt{3}}

x_{3, 4} = (- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i) x_{1}; x_{5, 6} = (- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i) x_{2}

and y_{n} = \frac{1}{x_{n}}

Commented by benjo last updated on 25/Dec/19

it mean the equation has 6 roots

sir ?

Commented by MJS last updated on 25/Dec/19

yes

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