Answers

2014-09-26T20:36:29+05:30
 \frac{x}{y} + \frac{y}{x} = -1
On taking cube both side of the equation .,
 ( \frac{x}{y} + \frac{y}{x}) ^{3} = (-1)^{3}  \\ or,  (\frac{x}{y}) ^{3} + ( \frac{y}{x} )^{3} +3. (  \frac{x}{y}) ^{2} .\frac{y}{x} +3 .\frac{x}{y}  . (\frac{y}{x} )^{2} =-1
 \frac{ x^{3} }{ y^{3} } + \frac{ y^{3} }{ x^{3} } +3( \frac{x}{y} + \frac{y}{x} ) =-1 \\ or, \frac{ x^{6}+ y^{6}  }{ x^{3}  y^{3} } +3(-1)=-1
or, x^{6} + y^{6} =2 x^{3} . y^{3}  \\ or,  (x^{3}) ^{2} + (y^{3})  x^{2} -2 x^{3} . y^{3} =0
or, ( x^{3} - y^{3} )^{2} =0 \\ so, x^{3} - y^{3} =0
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mark plse
thnx
please..
gd
this solution is also good. contnue giving good answers.
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2014-09-26T20:42:42+05:30

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\frac{x}{y}+\frac{y}{x} = -1 \\ \\ \frac{x^2+y^2}{xy} = - 1 \\ \\ x^2+y^2 = -xy \\ \\ x^2+y^2+xy = 0 \\ \\ x^3-y^3 = (x-y)(x^2+xy+y^2) = (x-y)* 0 = 0 \\

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