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Fractions
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June 14, 2021 9:53 am
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(@armor)
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x,y are positive real numbers and   \frac{1}{x}-\frac{1}{y}-\frac{1}{x+y}=0,  what is the value of  (\frac{y}{x})^3+(\frac{x}{y})^3?

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June 14, 2021 9:53 am
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

\small \frac{1}{x}-\frac{1}{y}-\frac{1}{x+y}=0\Rightarrow \frac{y-x}{xy}=\frac{1}{x+y}\Rightarrow \frac{y}{x}-\frac{x}{y}=1,   thus

\small \frac{y}{x}+\frac{x}{y}=\sqrt{(\frac{y}{x}-\frac{x}{y})^2+4\frac{y}{x}\frac{x}{y}}=\sqrt{5}      Therefore

\small (\frac{y}{x})^3+(\frac{x}{y})^3=(\frac{y}{x}+\frac{x}{y})(\frac{y^2}{x^2}-\frac{y}{x}\frac{x}{y}+\frac{x^2}{y^2})

\small =(\frac{y}{x}+\frac{x}{y})[(\frac{y}{x}+\frac{x}{y})^2-3\frac{y}{x}\frac{x}{y}]=\sqrt{5}(5-3)=2\sqrt{5}.

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