a,b,c and d are positive integers. Quantity I: `'a':-((a+d)^(2)-(a-d)^(2))/(8ad(a+d)^(2))=1` Quantity II: `'b'((b+d)^(3)-(b-d)^(3))/((d^(2)+3b^(2)))=(1)/(8d)` Quantity III: `'c':-(sqrt(c+d)+sqrt(c-d))/(sqrt(c+d)-sqrt(c-d))`

Options

A. (@,&)
B. (#,@)
C. ($,@)
D. (&,@)

Explanation

To solve the problem, we need to analyze each quantity step by step.

### Step 1: Analyze Quantity I
We start with the equation given in Quantity I:

\[ -a \cdot \frac{(a + d)^2 - (a - d)^2}{8ad(a + d)^2} = 1 \]

1. **Expand the squares**:
\[ (a + d)^2 = a^2 + 2ad + d^2 \]
\[ (a - d)^2 = a^2 - 2ad + d^2 \]

2. **Subtract the two expansions**:
\[ (a + d)^2 - (a - d)^2 = (a^2 + 2ad + d^2) - (a^2 - 2ad + d^2) = 4ad \]

3. **Substitute back into the equation**:
\[ -a \cdot \frac{4ad}{8ad(a + d)^2} = 1 \]

4. **Simplify the fraction**:
\[ -a \cdot \frac{4}{8(a + d)^2} = 1 \implies -\frac{a}{2(a + d)^2} = 1 \]

5. **Rearranging gives**:
\[ a = -2(a + d)^2 \]
Since \(a\) and \(d\) are positive integers, this equation does not hold true. Thus, Quantity I does not yield a valid positive integer solution.

### Step 2: Analyze Quantity II
Now, we analyze Quantity II:

\[ b \cdot \frac{(b + d)^3 - (b - d)^3}{d^2 + 3b^2} = \frac{1}{8d} \]

1. **Expand the cubes**:
\[ (b + d)^3 = b^3 + 3b^2d + 3bd^2 + d^3 \]
\[ (b - d)^3 = b^3 - 3b^2d + 3bd^2 - d^3 \]

2. **Subtract the two expansions**:
\[ (b + d)^3 - (b - d)^3 = (b^3 + 3b^2d + 3bd^2 + d^3) - (b^3 - 3b^2d + 3bd^2 - d^3) = 6b^2d + 2d^3 \]

3. **Substituting back**:
\[ b \cdot \frac{6b^2d + 2d^3}{d^2 + 3b^2} = \frac{1}{8d} \]

4. **Cross-multiplying**:
\[ 8d \cdot b(6b^2d + 2d^3) = d^2 + 3b^2 \]

5. **This gives a relationship between \(b\) and \(d\)**, but solving for \(b\) explicitly is complex without specific values for \(d\).

### Step 3: Analyze Quantity III
Now we analyze Quantity III:

\[ -c \cdot \frac{\sqrt{c + d} + \sqrt{c - d}}{\sqrt{c + d} - \sqrt{c - d}} = 2 \]

1. **Rationalize the denominator**:
\[ \frac{\sqrt{c + d} + \sqrt{c - d}}{\sqrt{c + d} - \sqrt{c - d}} \cdot \frac{\sqrt{c + d} + \sqrt{c - d}}{\sqrt{c + d} + \sqrt{c - d}} = \frac{(c + d) - (c - d)}{(\sqrt{c + d} - \sqrt{c - d})^2} = \frac{2d}{(\sqrt{c + d} - \sqrt{c - d})^2} \]

2. **Substituting back**:
\[ -c \cdot \frac{2d}{(\sqrt{c + d} - \sqrt{c - d})^2} = 2 \]

3. **Rearranging gives**:
\[ c = -\frac{2(\sqrt{c + d} - \sqrt{c - d})^2}{2d} \]

This equation can also be complex to solve directly for \(c\).

### Conclusion
From the analysis, we find:

- **Quantity I** does not yield a valid solution.
- **Quantity II** and **Quantity III** are complex but can yield positive integer solutions depending on \(d\).

### Final Comparison
Since Quantity I does not yield a valid positive integer solution, we can conclude that:

- **Quantity III is greater than Quantity II** based on the derived relationships, but we cannot definitively compare Quantity I with the others.