1. **State the problem:** We are given the function $$f(x) = 1 - 2kx + kx^2 - x^3$$ and asked to find the range of values for the parameter $$k$$ such that $$f(x)$$ is decreasing for all real $$x$$. 2. **Recall decreasing condition:** A function is decreasing on its entire domain if its derivative $$f'(x) \leq 0$$ for all $$x$$. 3. **Find the derivative:** Differentiate $$f(x)$$ with respect to $$x$$: $$ f'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(2kx) + \frac{d}{dx}(kx^2) - \frac{d}{dx}(x^3) = 0 - 2k + 2kx - 3x^2 $$ So, $$ f'(x) = -2k + 2kx - 3x^2 = 2kx - 2k - 3x^2 $$ 4. **Rewrite the derivative:** $$ f'(x) = -3x^2 + 2kx - 2k $$ 5. **Condition for decreasing:** For $$f(x)$$ to be decreasing everywhere, we must have: $$ f'(x) \leq 0 \quad \text{for all } x \in \mathbb{R} $$ 6. **Analyze the quadratic polynomial in $$x$$:** Consider $$ g(x) = f'(x) = -3x^2 + 2kx - 2k $$ This is a quadratic with leading coefficient $$a = -3 < 0$$, so it opens downward. 7. **Maximum value condition:** Since the parabola opens downwards and we want $$g(x) \leq 0$$ everywhere, its maximum value must be $$\leq 0$$. 8. **Find the vertex:** The vertex of $$g(x)$$ is at $$ x_v = -\frac{b}{2a} = -\frac{2k}{2 \times (-3)} = \frac{k}{3} $$ 9. **Evaluate $$g(x)$$ at vertex:** $$ g(x_v) = -3 \left(\frac{k}{3}\right)^2 + 2k \times \frac{k}{3} - 2k = -3 \times \frac{k^2}{9} + \frac{2k^2}{3} - 2k = -\frac{k^2}{3} + \frac{2k^2}{3} - 2k = \frac{k^2}{3} - 2k $$ 10. **Set max value $$\leq 0$$:** $$ \frac{k^2}{3} - 2k \leq 0 $$ Multiply both sides by 3 (positive, so inequality direction stays same): $$ k^2 - 6k \leq 0 $$ 11. **Factor inequality:** $$ k(k - 6) \leq 0 $$ 12. **Solve the inequality:** This is true when $$k$$ lies between 0 and 6 inclusive: $$ 0 \leq k \leq 6 $$ 13. **Final answer:** The function $$$f(x)$$ is decreasing everywhere if and only if: $$ 0 \leq k \leq 6 $$ So the correct option is **B) 0 ≤ k ≤ 6**.