If a, b, c are all positive integers, then the minimum value of the expression `((a+b+c)(b^(2)+b+1)(c^(2)+c+1))/(abc)` is :

Options

A. 3
B. 9
C. 27
D. 1

Answer

To find the minimum value of the expression

\[ \frac{(a+b+c)(b^2+b+1)(c^2+c+1)}{abc} \]

where \(a\), \(b\), and \(c\) are all positive integers, we can follow these steps:

### Step 1: Substitute the minimum values for \(a\), \(b\), and \(c\)

Since \(a\), \(b\), and \(c\) are positive integers, the smallest value we can assign to each of them is 1. Thus, we set:

\[ a = 1, \quad b = 1, \quad c = 1 \]

### Step 2: Calculate \(a + b + c\)

Now, we calculate:

\[ a + b + c = 1 + 1 + 1 = 3 \]

### Step 3: Calculate \(b^2 + b + 1\)

Next, we calculate:

\[ b^2 + b + 1 = 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \]

### Step 4: Calculate \(c^2 + c + 1\)

Similarly, we calculate:

\[ c^2 + c + 1 = 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \]

### Step 5: Substitute these values into the expression

Now we substitute these values back into the expression:

\[ \frac{(3)(3)(3)}{(1)(1)(1)} = \frac{27}{1} = 27 \]

### Step 6: Conclusion

Thus, the minimum value of the expression is:

\[ \boxed{27} \]