1. **Problem:** Given that $$\sum_{r=1}^k r^3 = \frac{k^2 (k+1)^2}{4}$$ and $$k$$ is a positive integer, find $$\sum_{r=3}^k (r-1)(r^2 + r + 1)$$ in terms of $$k$$. 2. **Step 1: Simplify the summand** Note that: $$(r-1)(r^2 + r + 1) = r^3 - 1$$ This is because: $$ (r-1)(r^2 + r + 1) = r^3 + r^2 + r - r^2 - r - 1 = r^3 - 1 $$ 3. **Step 2: Express the sum using this simplification** $$ \sum_{r=3}^k (r-1)(r^2 + r + 1) = \sum_{r=3}^k (r^3 - 1) = \sum_{r=3}^k r^3 - \sum_{r=3}^k 1 $$ 4. **Step 3: Use the given formula for sum of cubes** We know: $$ \sum_{r=1}^k r^3 = \frac{k^2 (k+1)^2}{4} $$ So, $$ \sum_{r=3}^k r^3 = \sum_{r=1}^k r^3 - \sum_{r=1}^2 r^3 = \frac{k^2 (k+1)^2}{4} - (1^3 + 2^3) = \frac{k^2 (k+1)^2}{4} - (1 + 8) = \frac{k^2 (k+1)^2}{4} - 9 $$ 5. **Step 4: Calculate the sum of constants** $$ \sum_{r=3}^k 1 = k - 2 $$ 6. **Step 5: Combine results** $$ \sum_{r=3}^k (r-1)(r^2 + r + 1) = \left( \frac{k^2 (k+1)^2}{4} - 9 \right) - (k - 2) = \frac{k^2 (k+1)^2}{4} - 9 - k + 2 = \frac{k^2 (k+1)^2}{4} - k - 7 $$ **Final answer:** $$ \sum_{r=3}^k (r-1)(r^2 + r + 1) = \frac{k^2 (k+1)^2}{4} - k - 7 $$