`N = 55^3 + 17^3 - 72^3`, then N is divisible by :

Options

A. 3 & 17
B. 40 & 11
C. 11 & 15
D. all of these

Answer

To solve the problem \( N = 55^3 + 17^3 - 72^3 \) and determine what \( N \) is divisible by, we can use the identity for the sum of cubes. The identity states that if \( a + b + c = 0 \), then:

\[ a^3 + b^3 + c^3 = 3abc \]

In our case, we can set:

- \( a = 55 \)
- \( b = 17 \)
- \( c = -72 \)

Now, we check if \( a + b + c = 0 \):

\[ 55 + 17 - 72 = 0 \]

Since this holds true, we can apply the identity:

\[ N = 55^3 + 17^3 - 72^3 = 3 \cdot 55 \cdot 17 \cdot (-72) \]

Next, we calculate \( 3 \cdot 55 \cdot 17 \cdot (-72) \):

1. Calculate \( 55 \cdot 17 \):
\[ 55 \cdot 17 = 935 \]

2. Calculate \( 935 \cdot (-72) \):
\[ 935 \cdot (-72) = -67260 \]

3. Now multiply by 3:
\[ N = 3 \cdot (-67260) = -201780 \]

Now, we need to find the divisors of \( N \). Since \( N \) is expressed as \( 3 \cdot 55 \cdot 17 \cdot (-72) \), we can find the prime factorization of \( 55 \) and \( 72 \):

- \( 55 = 5 \cdot 11 \)
- \( 72 = 8 \cdot 9 = 2^3 \cdot 3^2 \)

Thus, the prime factorization of \( N \) is:

\[ N = 3^3 \cdot 5^1 \cdot 11^1 \cdot 2^3 \]

Now we can conclude that \( N \) is divisible by:

- \( 3 \)
- \( 5 \)
- \( 11 \)
- \( 2 \)

To summarize, \( N \) is divisible by \( 3, 5, 11, 2 \), and any combinations of these.