1. Given the equation $x - 3 = -\frac{1}{x}$, our goal is to find the value of $x^3 + \frac{1}{x^3}$. 2. First, rewrite the equation: $$x - 3 = -\frac{1}{x} \implies x + \frac{1}{x} = 3$$ This step adds $\frac{1}{x}$ to both sides. 3. Let $y = x + \frac{1}{x}$, so we have $y = 3$. 4. We know the identity: $$\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right)$$ 5. Substitute $y = 3$ into the identity: $$y^3 = x^3 + \frac{1}{x^3} + 3y$$ 6. Rearranging to solve for $x^3 + \frac{1}{x^3}$: $$x^3 + \frac{1}{x^3} = y^3 - 3y$$ 7. Substitute $y=3$: $$x^3 + \frac{1}{x^3} = 3^3 - 3 \cdot 3 = 27 - 9 = 18$$ Final Answer: $x^3 + \frac{1}{x^3} = 18$