We are given the sum:

$\sum_{i=1}^{n} \left( i^3 - (i-1)^3 \right)$

Step 1: Expand the cube term

Expanding $(i-1)^3$ using the binomial theorem:

$(i-1)^3 = i^3 - 3i^2 + 3i - 1$

Thus, the given sum simplifies to:

$\sum_{i=1}^{n} \left( i^3 - (i^3 - 3i^2 + 3i - 1) \right)$

$= \sum_{i=1}^{n} \left( 3i^2 - 3i + 1 \right)$

Step 2: Write out terms explicitly

Let’s expand a few terms to observe the pattern:

$(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \dots + (n^3 - (n-1)^3)$

Most terms in this telescoping sum cancel out, leaving:

$n^3 - 0^3 = n^3$

Conclusion

The simplified form of the given sum is:

$\sum_{i=1}^{n} (i^3 - (i-1)^3) = n^3$

Would you like a deeper breakdown or have any questions? Here are some related questions you might consider:

1. How does the telescoping property help simplify summations?
2. Can we generalize this technique to higher powers?
3. What is the significance of expanding the terms before summing?
4. How can we prove this result using mathematical induction?
5. How does this relate to the sum of cubes formula?

Tip: When dealing with summations, always consider expanding terms to identify potential cancellations.