The number of triplets (a, b, c) of positive integers satisfying the equation `|(a^(3)+1,a^(2)b,a^(2)c),(ab^(2),b^(3)+1,b^(2)c),(ac^(2),bc^(2),c^(3)+1)|=30` is equal to

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem of finding the number of triplets \((a, b, c)\) of positive integers satisfying the equation

\[ |(a^3 + 1, a^2 b, a^2 c), (ab^2, b^3 + 1, b^2 c), (ac^2, bc^2, c^3 + 1)| = 30, \]

we will follow these steps:

### Step 1: Simplify the Determinant
We start with the given determinant and manipulate it. We can divide the first row by \(a\), the second row by \(b\), and the third row by \(c\):

\[ |(a^3 + 1, a^2 b, a^2 c), (ab^2, b^3 + 1, b^2 c), (ac^2, bc^2, c^3 + 1)|. \]

### Step 2: Rewriting the Determinant
After dividing, we can rewrite the determinant as:

\[ |(a^2, b^2 + \frac{1}{b}, c^2), (b^2, b^2 + 1, c^2), (c^2, b^2, c^2 + \frac{1}{c})|. \]

### Step 3: Multiply Rows
Next, we multiply the first row by \(a\), the second row by \(b\), and the third row by \(c\):

\[ |(a^3 + 1, a^2 b, a^2 c), (ab^2, b^3 + 1, b^2 c), (ac^2, bc^2, c^3 + 1)|. \]

### Step 4: Add Rows
Now, we add the three rows together. This gives us:

\[ |(a^3 + b^3 + c^3 + 3, a^2 b + b^2 + c^2, a^2 c + b^2 c + c^3 + 1)|. \]

### Step 5: Setting Up the Equation
The determinant simplifies to:

\[ a^3 + b^3 + c^3 + 1 = 30. \]

Thus, we have:

\[ a^3 + b^3 + c^3 = 29. \]

### Step 6: Finding Positive Integer Solutions
Now we need to find the positive integer solutions to the equation \(a^3 + b^3 + c^3 = 29\).

### Step 7: Testing Possible Values
We can test combinations of \(a\), \(b\), and \(c\):

1. **Case 1**: \(a = 3, b = 1, c = 1\) gives \(27 + 1 + 1 = 29\).
2. **Case 2**: \(a = 1, b = 3, c = 1\) gives \(1 + 27 + 1 = 29\).
3. **Case 3**: \(a = 1, b = 1, c = 3\) gives \(1 + 1 + 27 = 29\).

These give us the triplets \((3, 1, 1)\), \((1, 3, 1)\), and \((1, 1, 3)\).

### Step 8: Counting Unique Triplets
The unique triplets that satisfy the equation are:

- \((3, 1, 1)\)
- \((1, 3, 1)\)
- \((1, 1, 3)\)

Since the order matters in triplets, we can permute these values. The total number of distinct permutations for the triplet \((3, 1, 1)\) is given by:

\[ \frac{3!}{2!} = 3. \]

### Final Count
Thus, we have 3 unique arrangements for each of the triplet combinations.

### Conclusion
The total number of triplets \((a, b, c)\) of positive integers satisfying the equation is:

\[ \boxed{3}. \]