Find all pairs of complex numbers (x,y) satisfying $xy=-1$ and $x^3-y^3=-1$ the problem asks me to find all pairs of complex numbers (x,y) satisfying $xy=-1$ and $x^3-y^3=-1$ simultaneously. I set the two equations to equal each other which yields
$x^3-xy-y^3=0$ but from this point on on im stuck on how i would solve for $x$ and $y$. any help would be appreciated thanks!
 A: You have
$(x^3)(-y^3)=-(-1)^3=1$
$x^3+(-y^3)=-1$
You are to derive a quadratic equation whose roots are $x^3,-y^3$.  Pick either root as $x^3$, take cube roots and derive the value of $y$ for each $x$ from the product relation.  Good luck!
A: It seems that anon gave some explanation of the suggestion I made.
Also, you do have questions using complex numbers, including some with $e^{m \pi i / n}.$
So, your original variable $x$ satisfies $x^6 + x^3 + 1 = 0.$ Note that we have $x^3 \neq 1,$ otherwise we would have $1+1+1=0,$ which is false. Next,
$$ 0 = (x^3 - 1)(x^6 + x^3 + 1) = x^9 - 1. $$ Notice that the comment by anon points out that the solutions $x$ are precisely the primitive 9th roots of unity. 
So far, $$ x^9 = 1, \; \; \mbox{but} \; \; \; x^3 \neq 1.  $$
Define $$  \omega = e^{2 \pi i / 9} = \cos \frac{2 \pi}{9} + i \, \sin \frac{2 \pi}{9} \; \; . $$
The six answers to the original problem are $$ x = \omega, \omega^2, \omega^4, \omega^5, \omega^7, \omega^8. $$
In each case we take
$$ y = - \bar{x}.  $$
To be specific, when $x = a + b i,$ we get $y = -a + bi.$
For example, when $x = \omega,$ we get
$$ y = - \cos \frac{2 \pi}{9} + i \, \sin \frac{2 \pi}{9} = e^{7 \pi i / 9} \; \; .  $$
