If p,q,r are such that q-p = r-q = 2 for every `p,q,r in I^+`, then the least possible value of (p+q)(q+r)(r+p) is :

Options

A. 192
B. 720
C. 129
D. data insufficient

Answer

To solve the problem, we need to find the least possible value of the expression \((p+q)(q+r)(r+p)\) given that \(q - p = r - q = 2\).

### Step 1: Express \(p\), \(q\), and \(r\) in terms of a single variable.
Since \(q - p = 2\) and \(r - q = 2\), we can express \(q\) and \(r\) in terms of \(p\):
- Let \(p = a\)
- Then, \(q = p + 2 = a + 2\)
- And \(r = q + 2 = (a + 2) + 2 = a + 4\)

### Step 2: Substitute \(p\), \(q\), and \(r\) into the expression.
Now we substitute \(p\), \(q\), and \(r\) into the expression \((p+q)(q+r)(r+p)\):
- \(p + q = a + (a + 2) = 2a + 2\)
- \(q + r = (a + 2) + (a + 4) = 2a + 6\)
- \(r + p = (a + 4) + a = 2a + 4\)

### Step 3: Write the expression in terms of \(a\).
Now we can write the expression as:
\[ (2a + 2)(2a + 6)(2a + 4) \]

### Step 4: Simplify the expression.
We can factor out 2 from each term:
\[ = 2 \cdot (a + 1)(2a + 6)(a + 2) \]
This simplifies to:
\[ = 8(a + 1)(a + 2)(a + 3) \]

### Step 5: Find the minimum value.
To find the minimum value of \((a + 1)(a + 2)(a + 3)\), we need to evaluate it at the smallest positive integer value for \(a\). The smallest positive integer is \(1\):
- When \(a = 1\):
\[ (1 + 1)(1 + 2)(1 + 3) = 2 \cdot 3 \cdot 4 = 24 \]

### Step 6: Calculate the final value.
Now, substituting back into our expression:
\[ 8 \cdot 24 = 192 \]

Thus, the least possible value of \((p+q)(q+r)(r+p)\) is **192**.