Question

If x/y+y/x=-1 ,find x^3-y^3??PLS BE FAST ...HVING XAMS...IM HVING A DOUBT IN TIS QUES...

Answer 1

$\frac{x}{y} + \frac{y}{x} = -1$
On taking cube both side of the equation .,
$( \frac{x}{y} + \frac{y}{x}) ^{3} = (-1)^{3} \\ or, (\frac{x}{y}) ^{3} + ( \frac{y}{x} )^{3} +3. ( \frac{x}{y}) ^{2} .\frac{y}{x} +3 .\frac{x}{y} . (\frac{y}{x} )^{2} =-1$
$\frac{ x^{3} }{ y^{3} } + \frac{ y^{3} }{ x^{3} } +3( \frac{x}{y} + \frac{y}{x} ) =-1 \\ or, \frac{ x^{6}+ y^{6} }{ x^{3} y^{3} } +3(-1)=-1$
$or, x^{6} + y^{6} =2 x^{3} . y^{3} \\ or, (x^{3}) ^{2} + (y^{3}) x^{2} -2 x^{3} . y^{3} =0$
$or, ( x^{3} - y^{3} )^{2} =0 \\ so, x^{3} - y^{3} =0$

Answer 2

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