`94^(3)-23^(3)-71^(3)` is atleast divisible by

Options

A. 71 and 23
B. 23 and 74
C. 71 and 94
D. 23 ,71 and 94

Answer

To solve the problem \( 94^3 - 23^3 - 71^3 \) and determine by what number it is at least divisible, we can use the properties of cubes and the difference of cubes formula.

### Step-by-step Solution:

1. **Identify the Expression**: We start with the expression \( 94^3 - 23^3 - 71^3 \).

2. **Rearranging the Expression**: We can rearrange the expression to group \( 94^3 \) with \( 23^3 + 71^3 \):
\[ 94^3 - (23^3 + 71^3) \]

3. **Using the Sum of Cubes Formula**: We know that \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, let \( a = 23 \) and \( b = 71 \):
\[ 23^3 + 71^3 = (23 + 71)(23^2 - 23 \cdot 71 + 71^2) \]
This simplifies to:
\[ 23^3 + 71^3 = 94(23^2 - 23 \cdot 71 + 71^2) \]

4. **Substituting Back**: Now we substitute back into our expression:
\[ 94^3 - (23^3 + 71^3) = 94^3 - 94(23^2 - 23 \cdot 71 + 71^2) \]

5. **Factoring Out 94**: We can factor out 94:
\[ = 94 \left( 94^2 - (23^2 - 23 \cdot 71 + 71^2) \right) \]

6. **Conclusion on Divisibility**: Since we have factored out 94, it shows that \( 94^3 - 23^3 - 71^3 \) is divisible by 94.

### Final Result:
Thus, \( 94^3 - 23^3 - 71^3 \) is at least divisible by **94**.