If x/y+y/x=-1 (where, x, y is not equal to zero), then find the value of x^3-y^3.

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x^{3} + y^{3} + z^{3} = 6 x y z

x + y + z \neq 0

x y z = x + y + z

(x - y)^{2} + (y - z)^{2} + (z - x)^{2}

If $\frac{x}{y} + \frac{y}{x} = - 1$ , where $x, y \neq 0$ then the value of $(x^{3} - y^{3})$ is
(a) $1$

(b) $- 1$

(d) $\frac{1}{2}$

\frac{x}{y} + \frac{y}{x} = - 1, (x, y \neq 0)

x^{3} - y^{3}

x^{3} + y^{3} + 1 = 3 x y,

x \neq y

x + y + 1.

\frac{x}{y} + \frac{y}{x} = - 1 (x, y \neq 0)

x^{3} - y^{3}

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