1. **State the problem:** Find all critical points of the function $$f(x) = x^3 - 6x^2 + 9x + a$$ where $$a \in \mathbb{R}$$. 2. **Recall the formula for critical points:** Critical points occur where the derivative $$f'(x)$$ is zero or undefined. Since $$f(x)$$ is a polynomial, $$f'(x)$$ is defined everywhere, so we only solve $$f'(x) = 0$$. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + a) = 3x^2 - 12x + 9$$ 4. **Set the derivative equal to zero:** $$3x^2 - 12x + 9 = 0$$ 5. **Simplify the equation:** Divide both sides by 3: $$x^2 - 4x + 3 = 0$$ 6. **Factor the quadratic:** $$(x - 3)(x - 1) = 0$$ 7. **Solve for $$x$$:** $$x = 3 \quad \text{or} \quad x = 1$$ 8. **Conclusion:** The critical points of $$f(x)$$ are at $$x = 1$$ and $$x = 3$$, independent of the value of $$a$$.