If a and b are positive integers such that `N=(a+ib)^(3)-107i` (where N is a natural number), then the value of a is equal to (where `i^(2)=-1`)

Options

A. 4
B. 5
C. 6
D. 9

Answer

To solve the problem, we need to find the value of \( a \) given that \( N = (a + ib)^3 - 107i \) is a natural number, where \( a \) and \( b \) are positive integers.

### Step-by-Step Solution:

1. **Expand the expression**:
We start by expanding \( (a + ib)^3 \):
\[ (a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 \]
Using \( i^2 = -1 \) and \( i^3 = -i \), we can rewrite this as:
\[ = a^3 + 3a^2(ib) + 3a(-b^2) + (-ib^3) \]
This simplifies to:
\[ = a^3 - 3ab^2 + i(3a^2b - b^3) \]

2. **Combine with -107i**:
Now we substitute this back into the equation for \( N \):
\[ N = (a^3 - 3ab^2) + i(3a^2b - b^3 - 107) \]
For \( N \) to be a natural number, the imaginary part must be zero:
\[ 3a^2b - b^3 - 107 = 0 \]

3. **Rearranging the equation**:
Rearranging gives:
\[ 3a^2b = b^3 + 107 \]
Dividing both sides by \( b \) (since \( b \) is positive):
\[ 3a^2 = b^2 + \frac{107}{b} \]

4. **Finding conditions for \( b \)**:
For \( 3a^2 \) to be an integer, \( \frac{107}{b} \) must also be an integer. Therefore, \( b \) must be a divisor of 107. The positive divisors of 107 are 1 and 107.

5. **Testing \( b = 1 \)**:
If \( b = 1 \):
\[ 3a^2 = 1^2 + \frac{107}{1} = 1 + 107 = 108 \]
Thus,
\[ 3a^2 = 108 \implies a^2 = \frac{108}{3} = 36 \implies a = 6 \]

6. **Testing \( b = 107 \)**:
If \( b = 107 \):
\[ 3a^2 = 107^2 + \frac{107}{107} = 11449 + 1 = 11450 \]
This gives:
\[ 3a^2 = 11450 \implies a^2 = \frac{11450}{3} \approx 3816.67 \]
Since \( a^2 \) must be an integer, this case is not valid.

7. **Conclusion**:
The only valid solution occurs when \( b = 1 \), giving us \( a = 6 \).

### Final Answer:
The value of \( a \) is \( \boxed{6} \).