Given that $x^2 + y^2 + xy = 0$, we want to find the value of $(\frac{x-y}{x})^{80} + (\frac{x-y}{y})^{80}$.

1. Problem Description

Given that $x^2 + y^2 + xy = 0$, we want to find the value of $(\frac{x-y}{x})^{80} + (\frac{x-y}{y})^{80}$.

2. Solution Steps

From the equation $x^2 + y^2 + xy = 0$, we can divide by $y^2$ (assuming $y \ne 0$):

$(\frac{x}{y})^2 + 1 + \frac{x}{y} = 0$

Let $t = \frac{x}{y}$. Then we have $t^2 + t + 1 = 0$.

Using the quadratic formula to solve for $t$,

$t = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$

So, $t = \frac{-1 + i\sqrt{3}}{2}$ or $t = \frac{-1 - i\sqrt{3}}{2}$. These are the two complex cube roots of unity, excluding

1. Let $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.

Thus, $\frac{x}{y} = \omega$ or $\frac{x}{y} = \omega^2$. In either case, $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

Case 1: $\frac{x}{y} = \omega$. Then $x = \omega y$.

$\frac{x-y}{x} = \frac{\omega y - y}{\omega y} = \frac{\omega - 1}{\omega} = 1 - \frac{1}{\omega} = 1 - \omega^2$.

Since $1 + \omega + \omega^2 = 0$, we have $1 + \omega = -\omega^2$, so $1 - \omega^2 = 1 - \frac{x^2}{y^2}$.

Also, since $1 + \omega + \omega^2 = 0$, we have $-\omega^2 = 1 + \omega$, so $1 - \omega^2 = 1 + (1 + \omega) = 2 + \omega$.

Then $\frac{x-y}{x} = 1 - \omega^2 = -\omega$.

$\frac{x-y}{y} = \frac{\omega y - y}{y} = \omega - 1$. Since $1 + \omega + \omega^2 = 0$, $\omega - 1 = -2 - \omega^2 = -2 - \omega^2$.

$\frac{x-y}{y} = \omega - 1 = -1 - \omega^2$. Since $\omega^2 + \omega + 1 = 0$, $\omega^2 = -1 - \omega$. Thus $\omega - 1 = \omega - 1$. Also $\omega - 1 = -1 - \omega^2$, which is equal to $-\omega^2 - 1 = \omega$.

$\frac{x-y}{x} = \frac{\omega y - y}{\omega y} = \frac{\omega - 1}{\omega} = 1 - \frac{1}{\omega} = 1 - \omega^2$.

$\frac{x-y}{y} = \frac{\omega y - y}{y} = \omega - 1$.

Since $\omega^3 = 1$, we have $\frac{1}{\omega} = \omega^2$. Then $\frac{x-y}{x} = 1 - \omega^2 = -\omega$.

Therefore, $(\frac{x-y}{x})^{80} = (-\omega)^{80} = \omega^{80} = \omega^{2}$ since $80 = 3(26) + 2$.

Since $\omega - 1 = \frac{-1 + i\sqrt{3}}{2} - 1 = \frac{-3 + i\sqrt{3}}{2}$,

$\frac{x-y}{y} = \omega - 1$.

Then $(\frac{x-y}{y})^{80} = (\omega - 1)^{80} = (-\omega^2 - 2)^{80} = (\frac{-3+i\sqrt{3}}{2})^{80}$. This is incorrect.

Going back to $x^2 + y^2 + xy = 0$. Multiply by $x-y$.

$(x-y)(x^2 + y^2 + xy) = x^3 - y^3 = 0$, thus $x^3 = y^3$.

Then $\frac{x}{y}$ is a cube root of unity, not equal to 1, as mentioned.

$\frac{x-y}{x} = 1 - \frac{y}{x}$ and $\frac{x-y}{y} = \frac{x}{y} - 1$.

If $x^3 = y^3$, then $(\frac{x}{y})^3 = 1$, which gives $\frac{x}{y} = \omega$ or $\omega^2$ (where $\omega$ is a complex cube root of unity).

Thus, $\frac{y}{x} = \omega^2$ or $\omega$.

$\frac{x-y}{x} = 1 - \omega^2 = -\omega$ or $1 - \omega = -\omega^2$.

$\frac{x-y}{y} = \omega - 1 = -\omega^2$ or $\omega^2 - 1 = -\omega$.

Thus, $(\frac{x-y}{x})^{80} = (-\omega)^{80} = \omega^{80} = (\omega^3)^{26} \omega^2 = \omega^2$ or $(-\omega^2)^{80} = \omega^{160} = (\omega^3)^{53} \omega = \omega$.

$(\frac{x-y}{y})^{80} = (-\omega^2)^{80} = \omega^{160} = (\omega^3)^{53} \omega = \omega$ or $(-\omega)^{80} = \omega^{80} = (\omega^3)^{26} \omega^2 = \omega^2$.

In either case, the sum is $\omega + \omega^2 = -1$.

3. Final Answer

-1