1. **State the problem:** Solve the equation $$\frac{n \times n \times n - n}{n^2} = 9$$ for $n$. 2. **Rewrite the expression:** The numerator is $n \times n \times n - n = n^3 - n$. 3. **Substitute into the equation:** $$\frac{n^3 - n}{n^2} = 9$$ 4. **Simplify the fraction:** $$\frac{n^3}{n^2} - \frac{n}{n^2} = n - \frac{1}{n}$$ 5. **Rewrite the equation:** $$n - \frac{1}{n} = 9$$ 6. **Multiply both sides by $n$ to clear the denominator:** $$n^2 - 1 = 9n$$ 7. **Rearrange to standard quadratic form:** $$n^2 - 9n - 1 = 0$$ 8. **Use the quadratic formula:** $$n = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{9 \pm \sqrt{81 + 4}}{2} = \frac{9 \pm \sqrt{85}}{2}$$ 9. **Final solutions:** $$n = \frac{9 + \sqrt{85}}{2} \quad \text{or} \quad n = \frac{9 - \sqrt{85}}{2}$$ These are the two values of $n$ that satisfy the original equation.