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Sum of Cubes of First $$n$$ Natural Numbers
## Formula
$$1^3+2^3+3^3+\cdots+n^3=\left[\frac{n(n+1)}2\right]^2$$
Therefore:
$$\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}2\right]^2$$
## Usage
- Used in summation problems involving cubes of consecutive natural numbers.
