1. **State the problem:** Solve the cubic equation $$n^3 - 3n - 2 = 0$$ to find the exact roots. 2. **Recall the formula and approach:** For cubic equations of the form $$n^3 + an^2 + bn + c = 0$$, one method is to try to find rational roots using the Rational Root Theorem, then factor and solve the remaining quadratic. 3. **Apply Rational Root Theorem:** Possible rational roots are factors of the constant term 2 divided by factors of the leading coefficient 1, i.e., $$\pm1, \pm2$$. 4. **Test roots:** - For $$n=1$$: $$1^3 - 3(1) - 2 = 1 - 3 - 2 = -4 \neq 0$$ - For $$n=-1$$: $$(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$$, so $$n = -1$$ is a root. 5. **Factor out $$n + 1$$:** Use polynomial division or synthetic division: $$n^3 - 3n - 2 = (n + 1)(\cancel{n^3} - n^2 + n - 2)$$ Performing division: $$\frac{n^3 - 3n - 2}{n + 1} = n^2 - n - 2$$ 6. **Solve quadratic factor:** $$n^2 - n - 2 = 0$$ Use quadratic formula: $$n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2}$$ 7. **Simplify roots:** $$n = \frac{1 \pm 3}{2}$$ - $$n = \frac{1 + 3}{2} = 2$$ - $$n = \frac{1 - 3}{2} = -1$$ 8. **List all roots:** From step 4 and 7, roots are $$n = -1$$ (double root) and $$n = 2$$. 9. **Check multiplicity:** Since $$n = -1$$ was a root of the cubic and also a root of the quadratic factor, it is a repeated root. **Final answer:** The exact roots of $$n^3 - 3n - 2 = 0$$ are $$\boxed{n = -1 \text{ (double root), } n = 2}$$.