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 \( (a + ib)^3 \)**:
\[ (a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 \]
Using \( i^2 = -1 \), we can simplify:
\[ = a^3 + 3a^2(ib) + 3a(-b^2) + (i^3b^3) = a^3 - 3ab^2 + i(3a^2b - b^3) \]
Thus, we can write:
\[ (a + ib)^3 = (a^3 - 3ab^2) + i(3a^2b - b^3) \]

2. **Substituting into \( N \)**:
\[ N = (a^3 - 3ab^2) + i(3a^2b - b^3) - 107i \]
This simplifies to:
\[ N = (a^3 - 3ab^2) + i(3a^2b - b^3 - 107) \]

3. **Separating real and imaginary parts**:
For \( N \) to be a natural number, the imaginary part must equal zero:
\[ 3a^2b - b^3 - 107 = 0 \]
Rearranging gives:
\[ 3a^2b - b^3 = 107 \]

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

5. **Finding suitable values for \( b \)**:
Since \( b \) must be a divisor of 107, the positive divisors of 107 are 1 and 107. We will check these values.

- **Case 1: \( b = 1 \)**:
\[ 3a^2 = 1^2 + \frac{107}{1} = 1 + 107 = 108 \]
\[ 3a^2 = 108 \implies a^2 = 36 \implies a = 6 \]

- **Case 2: \( b = 107 \)**:
\[ 3a^2 = 107^2 + \frac{107}{107} = 11449 + 1 = 11450 \]
\[ 3a^2 = 11450 \implies a^2 = \frac{11450}{3} \text{ (not an integer)} \]

6. **Conclusion**:
The only valid solution is when \( b = 1 \) and \( a = 6 \). Thus, the value of \( a \) is:
\[ \boxed{6} \]