Question

x/y+y/z=-1 find x^3-y^3

Answer 1

Answer will be 0

I think you have written wrong question
there should be X in place of Z

$\frac{x}{y} + \frac{y}{x} = - 1$
=>
${x}^{2} + {y}^{2} = - xy$


${x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)$
=
$(x - y)( - yx + xy)$
= 0