Find the sum of cubes of first 10 natural number .

Answer

To find the sum of the cubes of the first 10 natural numbers, we can use the formula for the sum of cubes:

\[ \text{Sum of cubes of the first } n \text{ natural numbers} = \left( \frac{n(n+1)}{2} \right)^2 \]

### Step-by-step Solution:

1. **Identify \( n \)**:
We are asked to find the sum of the cubes of the first 10 natural numbers, so we set \( n = 10 \).

2. **Apply the formula**:
Substitute \( n \) into the formula:
\[ \text{Sum} = \left( \frac{10(10+1)}{2} \right)^2 \]

3. **Calculate \( n(n+1) \)**:
Calculate \( 10 + 1 \):
\[ 10 + 1 = 11 \]
Now multiply:
\[ 10 \times 11 = 110 \]

4. **Divide by 2**:
Now divide by 2:
\[ \frac{110}{2} = 55 \]

5. **Square the result**:
Now square the result from the previous step:
\[ 55^2 = 3025 \]

6. **Final Answer**:
Therefore, the sum of the cubes of the first 10 natural numbers is:
\[ \text{Sum} = 3025 \]