An algebra problem by Manish Mayank

$\frac { x }{ y } +\frac { y }{ x } =-1\\ xy\left[ \frac { x }{ y } +\frac { y }{ x } \right] =-1\quad (xy)\\ { x }^{ 2 }+{ y }^{ 2 }=-xy\\ { x }^{ 2 }+xy+{ y }^{ 2 }=0\\ \\ using\quad quadratic\quad equation,\\ \\ x=\frac { { -y }\pm \sqrt { { y }^{ 2 }-4{ y }^{ 2 } } }{ 2 } \\ x=\frac { { -y }\pm y\sqrt { 3 } i }{ 2 } \\ \\ substitute\quad to\quad the\quad original\quad equation,\\ \frac { x }{ y } +\frac { y }{ x } =-1\\ \frac { { -y }+y\sqrt { 3 } i }{ 2y } +\frac { 2y }{ { -y }+y\sqrt { 3 } i } =-1\\ 2y\left( { -y }+y\sqrt { 3 } i \right) \left[ \frac { { -y }+y\sqrt { 3 } i }{ 2y } +\frac { 2y }{ { -y }+y\sqrt { 3 } i } \right] =(-1)\quad 2y\left( { -y }+y\sqrt { 3 } i \right) \\ simplify\quad the\quad equation,\quad we\quad can\quad get\quad the\quad value\quad of\quad y,\\ y=\frac { \sqrt { 3 } i+1 }{ 2 } \\ then\quad substitute\quad the\quad value\quad of\quad y\quad to\quad the\quad equation\quad of\quad the\quad x,\\ x=\frac { { -y }+y\sqrt { 3 } i }{ 2 } \\ x=\frac { -\frac { \sqrt { 3 } i+1 }{ 2 } +\frac { \sqrt { 3 } i+1 }{ 2 } (\sqrt { 3 } i) }{ 2 } \\ x=-1\\ \\ therefore\quad the\quad value\quad of\quad x\quad and\quad y\quad is\quad -1\quad \& \quad \frac { \sqrt { 3 } i+1 }{ 2 } \\ substitute\quad the\quad values\quad to\quad the\quad equation\quad { x }^{ 3 }-{ y }^{ 3 }\quad =\quad ?\\ { (-1) }^{ 3 }-{ \left( \frac { \sqrt { 3 } i+1 }{ 2 } \right) }^{ 3 }=\quad 0\\ \\ the\quad answer\quad is\quad 0.$

There is a short and easy solution to this without using the quadratic formula and complex numbers.

We can use the fact that $x^3 - y^3 = (x-y)(x^2+y^2-xy)$

From the question $\frac{x}{y} + \frac{y}{x} = -1$ , we get that

$x^2 +y^2 = -xy \Rightarrow x^2 +y^2 -xy = 0$

Substituting this in our identity above , we get $x^3-y^3 = 0$