Question

if x/y+y/x=-1, find the value of x^3-y^3

Answer 1

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$= > \it \frac{x}{y} + \frac{y}{x} = - 1 \\ \\ \bf \underline{squaring \: \: both \: \: sides \: \: we \: \: get} \\ \\ = > \it \: ( \frac{x}{y} + \frac{y}{x})^{2} = {( - 1)}^{2} \\ \\ = > \it\frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + 2 \frac{x}{y} . \frac{y}{x} = 1 \\ \\ = > \it \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + 2 = 1 \\ \\ = > \it \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } = 1 - 2 = - 1 \\ \\ = > \bf \: \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + 1 = \underline{0 }\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \underline{ \: \: now \: \: } \\ \\ = > \it \frac{ {x}^{3} }{ {y}^{3} } - \frac{ {y}^{3} }{ {x}^{3} } \\ \\ \it = ( \frac{x}{y} - \frac{y}{x} )( \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + \frac{x}{y} \times \frac{y}{x} ) \\ \\ \it = ( \frac{x}{y} - \frac{y}{x} )( \frac{ {x}^{2} }{ {y}^{2} } + \frac{ {y}^{2} }{ {x}^{2} } + 1) \\ \\ = \it \: ( \frac{x}{y} - \frac{y}{x} ) \times 0 \\ \\ = \bf \underline{0}$
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Answer 2

Hi,   Please see the attached file!    Thanks!