x^3 +y^3 = 35 (1/x)+(1/y) =(5/6) solve for all possible values of x and y


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Question Number 204421 by necx122 last updated on 17/Feb/24

$x^{3} + y^{3} = 35$ $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$ $s o l v e f o r a l l p o s s i b l e v a l u e s o f x a n d y$
	Answered by Rasheed.Sindhi last updated on 17/Feb/24

	$x^{3} + y^{3} = 35$ $\Rightarrow \frac{1}{x^{3}} + \frac{1}{y^{3}} = \frac{35}{x^{3} y^{3}}$ $\Rightarrow {(\frac{1}{x} + \frac{1}{y})}^{3} - 3 (\frac{1}{x}) (\frac{1}{y}) (\frac{1}{x} + \frac{1}{y}) = \frac{35}{x^{3} y^{3}}$ ${(\frac{5}{6})}^{3} - \frac{3}{x y} (\frac{5}{6}) = \frac{35}{x^{3} y^{3}}$ $\frac{125}{216} - \frac{5}{2 x y} = \frac{35}{x^{3} y^{3}}$ $\frac{125 x y - 540}{216 x y} = \frac{35}{x^{3} y^{3}}$ $\frac{125 x y - 540}{216} = \frac{35}{x^{2} y^{2}}$ $125 x^{3} y^{3} - 540 x^{2} y^{2} = 35 \times 216$ $25 {(x y)}^{3} - 108 {(x y)}^{2} - 1512 = 0$ $(x y - 6) (25 {(x y)}^{2} + 42 x + 252) = 0$ $x y = 6, \frac{- 42 \pm \sqrt{42^{2} - 4 (25) (252)}}{50}$ $x y = 6, \frac{- 21 \pm 3 i \sqrt{651}}{25}$ $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$ $\frac{x + y}{x y} = \frac{5}{6}$ $\frac{x + y}{6} = \frac{5}{6}$ $x + y = 5$ $x y = 6 \land x + y = 5 \Rightarrow (x, y) = (2, 3), (3, 2)$
	Commented by necx122 last updated on 17/Feb/24

	$W o w! T h i s i s l o v e l y a n d d e t a i l e d . T h a n k y o u$ $f o r a l w a y s c l e a r i n g m y d o u b t .$
	Answered by Rasheed.Sindhi last updated on 17/Feb/24

	$\frac{1}{x} + \frac{1}{y} = \frac{5}{6} \Rightarrow \frac{x + y}{x y} = \frac{5}{6} \Rightarrow x y = \frac{6 (x + y)}{5}$ $x^{3} + y^{3} = 35 \Rightarrow {(x + y)}^{3} - 3 x y (x + y) = 35$ $\Rightarrow {(x + y)}^{3} - 3 (\frac{6 (x + y)}{5}) (x + y) = 35$ $5 {(x + y)}^{3} - 18 {(x + y)}^{2} - 175 = 0$ $5 t^{3} - 18 t^{2} - 175 = 0; [t = x + y]$ $(t - 5) (5 t^{2} + 7 t + 35) = 0$ $t = 5, \frac{- 7 \pm i \sqrt{651}}{10}$ $x + y = 5, \frac{- 7 \pm i \sqrt{651}}{10}$ $x y = \frac{6 (x + y)}{5} = \frac{6 (5)}{5}, \frac{6 (\frac{- 7 \pm i \sqrt{651}}{10})}{5}$ $x y = 6, \frac{3 (- 7 \pm i \sqrt{651})}{25}$ $x + y = 5 \land x y = 6 \Rightarrow (x, y) = (2, 5), (5, 2)$ $x + y = \frac{- 7 \pm i \sqrt{651}}{10} \land x y = \frac{3 (- 7 \pm i \sqrt{651})}{25}$ $y = \frac{- 7 \pm i \sqrt{651}}{10} - x$ $\land x (\frac{- 7 \pm i \sqrt{651}}{10} - x) = \frac{3 (- 7 \pm i \sqrt{651})}{25}$ $\frac{- 7 \pm i \sqrt{651}}{10} x - x^{2} = \frac{3 (- 7 \pm i \sqrt{651})}{25}$ $x^{2} - \frac{- 7 \pm i \sqrt{651}}{10} x + \frac{3 (- 7 \pm i \sqrt{651})}{25} = 0$ $C o n t i n u e$
	Answered by Frix last updated on 17/Feb/24

	$x^{3} + y^{3} = 35$ $\frac{1}{x} + \frac{1}{y} = \frac{5}{6} \Leftrightarrow 5 x y = 6 (x + y)$ $x = u - v \land y = u + v$ $2 u^{3} + 6 {u v}^{2} = 35$ $5 u^{2} - 5 v^{2} = 12 u$ $\Rightarrow$ $\frac{35 - 2 u^{3}}{6 u} = \frac{5 u^{2} - 12 u}{5}$ $u^{3} - \frac{9 u^{2}}{5} - \frac{35}{8} = 0$ $u = \frac{5}{2} \Rightarrow v = \frac{1}{2}$ $u = - \frac{7}{20} \pm \frac{\sqrt{651}}{20} i \Rightarrow v = \frac{\sqrt{- 133 + 10 \sqrt{6433}}}{20} \mp \frac{\sqrt{133 + 10 \sqrt{6433}}}{20} i$ $\Rightarrow x, y$
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